CSP考前复习

前言

因为loceaner太菜了，他什么东西都不会

所以他打算学一个东西就记录一下

不过因为他很菜，所以他不会写原理……

而且，他希望在2019CSP之前不会断更

就酱紫，就是写给他自己的……因为他太菜了

基础算法

小技巧

\[\sum_{i = 0}^{x}C(x, i)* C(y, i) = C(x + y, x)
\]

使用负数下标

如何使用负数下标呢？

让数组前面有东西

int y[100];

int *z = y + 50;

这样的话调用\(z[-50]\)就变成了调用\(y[0]\)

z[-50] = y[0];

然后这样就可以实现调用啦~

其实还有一个更暴力的方法：用\(map\)

\(map\)是\(\log n\)的\(map\)

\(unordered\_map\)是\(O(1)\)的\(map\)（到\(c++11\)才会有）

二维前缀和

//知识点:二维前缀和

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 1000;

int n, m;

int a[N][N], b[N][N];

int main() {

	n = read(), m = read();

	for(int i = 1; i <= n; i++) {

		for(int j = 1; j <= n; j++) {

			a[i][j] = read();

			b[i][j] = a[i][j] + b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];

		}

	}

	for(int i, u1, v1, u2, v2; i <= m; i++) {

		u1 = read(), v1 = read(), u2 = read(), v2 = read();

		cout << b[u2][v2] - b[u1 - 1][v2] - b[u2][v1 - 1] + b[u1 - 1][v1 - 1] << '\n';

	}

	return 0;

}

三分法

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

const double eps = 1e-6;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n;

double l, r, a[A];

double f(double x) {

	double ans = 0;

	for(int i = 1; i <= n + 1; i++) ans = ans * x + a[i];

	return ans;

}

int main() {

	n = read(); scanf("%lf%lf", &l, &r);

	for(int i = 1; i <= n + 1; i++) scanf("%lf", &a[i]);

	while(r - l >= eps) {

		double qwq = (r - l) / 3.0, lmid = l + qwq, rmid = r - qwq;

		if(f(lmid) > f(rmid)) r = rmid;

		else l = lmid;

	}

	printf("%.5lf", l);

	return 0;

}

二维差分

//知识点:

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 1e3 + 11;

int n, m, a[N][N];

int main() {

	n = read(), m = read();

	for(int i = 1, u1, u2, v1, v2; i <= m; i++) {

		u1 = read(), v1 = read(), u2 = read(), v2 = read();

		a[u1][v1] += 1;

		a[u2 + 1][v2 + 1] += 1;

		a[u2 + 1][v1] -= 1;

		a[u1][v2 + 1] -= 1;

	}

	//C[x1][y1] += x ,  C[x2 + 1][y2 + 1] += x ,  C[x1][y2 + 1] -= x , C[x2 + 1][y1] -= x;

	for(int i = 1; i <= n; i++) {

		for(int j = 1; j <= n; j++) {

			a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];;

		}

	}

	for(int i = 1; i <= n; i++) {

		for(int j = 1; j <= n; j++) {

			cout << a[i][j] << ' ';

		}

		cout << '\n';

	}

	return 0;

}

归并排序

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

int n, m, a[A], b[A];

void solve(int l, int r) {

	if(l == r) return;

	int mid = (l + r) >> 1;

	solve(l, mid), solve(mid + 1, r);

	int i = l, j = mid + 1, k = l;

	while(i <= mid && j <= r) {

		if(a[i] <= a[j]) b[k++] = a[i++];

		else b[k++] = a[j++];

	}

	while(i <= mid) b[k++] = a[i++];

	while(j <= r) b[k++] = a[j++];

	for(int i = l; i <= r; i++) a[i] = b[i];

}

int main() {

//	freopen("a.in", "r", stdin);

	n = read();

	for(int i = 1; i <= n; i++) a[i] = read();

	solve(1, n);

	for(int i = 1; i <= n; i++) cout << a[i] << ' ';

	return 0;

}

排序不等式

给定3组数\(a, b, c\)

\(a[1]\)$a[n]$,$b[1]$\(b[n]\),\(c[1]\)~\(c[n]\)

其中\(c[1]\)$c[n]$是$b[1]$\(b[n]\)的乱序排列

\(a[1]*b[n]+a[2]*b[n-1]+...<=a[1]*c[1]+a[2]*c[2]+...<=a[1]*b[1]+a[2]*b[2]+...\)

即:逆序和 <= 乱序和 <= 正序和

关于\(long\ long\)

在日常的题目中，一定要看好数据范围，如果会爆\(int\)的话，不要忘记开\(long long\)（无数次被坑！！）

高精度模板

最让人烦的就是高精度了，某些题并不难，但是要写高精。。烦\(\color{white}{（ps:板子来自lfd）}\)

namespace BigInteger {

	struct Big_integer {

		int d[10005], len;

		void clean() {while(len > 1 and !d[len - 1]) len--;}

		Big_integer() {memset(d, 0, sizeof d);len = 1;}

		Big_integer(int num) {*this = num;}

		Big_integer operator = (const char* num) {

			memset(d, 0, sizeof d);

			len = strlen(num);

			for (int i = 0; i < len; i++) d[i] = num[len - 1 - i] - '0';

			clean();

			return *this;

		}

		Big_integer operator = (int num) {

			char s[10005];

			sprintf(s, "%d", num);

			*this = s;

			return *this;

		}

		Big_integer operator * (const Big_integer &b) const {

			int i, j;

			Big_integer c;

			c.len = len + b.len;

			for (j = 0; j < b.len; j++)

				for (i = 0; i < len; i++)

					c.d[i + j] += d[i] * b.d[j];

			for (i = 0; i < c.len - 1; i++) c.d[i + 1] += c.d[i] / 10, c.d[i] %= 10;

			c.clean();

			return c;

		}

		Big_integer operator / (const int &b) {

			int i, j, a = 0;

			Big_integer c = *this;

			for (i = len - 1; i >= 0; i--) {

				a = a * 10 + d[i];

				for (j = 0; j < 10; j++) if (a < b * (j + 1)) break;

				c.d[i] = j;

				a = a - b * j;

			}

			c.clean();

			return c;

		}

		bool operator < (const Big_integer &b) const {

			if (len != b.len) return len < b.len;

			for (int i = len - 1; i >= 0; i--)

				if (d[i] != b.d[i])

					return d[i] < b.d[i];

			return false;

		}

		string str() const {

			char s[10005];

			for (int i = 0; i < len; i++) s[len - 1 - i] = d[i] + '0';

			return s;

		}

	};

	istream& operator >> (istream& in, Big_integer &x) {

		string s;

		in >> s;

		x = s.c_str();

		return in;

	}

	ostream& operator << (ostream& out, const Big_integer &x) {

		out << x.str();

		return out;

	}

}

using namespace BigInteger;

数据结构

树状数组

单点修改，区间查询

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define lowbit(x) (x & (-x))

using namespace std;

const int A = 5e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, c[B];

void update(int x, int k) {

	while(x <= n) {

		c[x] += k;

		x += lowbit(x);

	}

}

int query(int x) {

	int ans = 0;

	while(x) {

		ans += c[x];

		x -= lowbit(x);

	}

	return ans;

}

int main() {

	n = read(), m = read();

	for(int i = 1, a; i <= n; i++) {

		a = read();

		update(i, a);

	}

	while(m--) {

		int opt = read(), x = read(), y = read();

		if(opt == 1) update(x, y);

		else cout << query(y) - query(x - 1) << '\n';

	}

	return 0;

}

区间修改，单点查询

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define lowbit(x) (x & (-x))

using namespace std;

const int A = 5e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, c[B], a[B];

void update(int x, int k) {

	while(x <= n) {

		c[x] += k;

		x += lowbit(x);

	}

}

int query(int x) {

	int ans = 0;

	while(x) {

		ans += c[x];

		x -= lowbit(x);

	}

	return ans;

}

int main() {

	n = read(), m = read();

	for(int i = 1; i <= n; i++) a[i] = read();

	while(m--) {

		int opt = read(), x = read(), y, k;

		if(opt == 1) y = read(), k = read(), update(x, k), update(y + 1, -k);

		else cout << a[x] + query(x) << '\n';

	}

	return 0;

}

线段树

区间修改，区间查询

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define int long long

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m;

namespace Seg {

	#define lson rt << 1

	#define rson rt << 1 | 1

	struct tree {

		int l, r, w, lazy;

	} t[A << 2];

	inline void pushup(int rt) { t[rt].w = t[lson].w + t[rson].w; }

	inline void pushdown(int rt) {

		t[lson].lazy += t[rt].lazy, t[rson].lazy += t[rt].lazy;

		t[lson].w += (t[lson].r - t[lson].l + 1) * t[rt].lazy;

		t[rson].w += (t[rson].r - t[rson].l + 1) * t[rt].lazy;

		t[rt].lazy = 0;

	}

	void build(int rt, int l, int r) {

		t[rt].l = l, t[rt].r = r;

		if(l == r) { t[rt].w = read(); return; }

		int mid = (l + r) >> 1;

		build(lson, l, mid), build(rson, mid + 1, r);

		pushup(rt); return;

	}

	void update(int rt, int l, int r, int val) {

		if(l <= t[rt].l && t[rt].r <= r) {

			t[rt].lazy += val;

			t[rt].w += (t[rt].r - t[rt].l + 1) * val;

			return;

		}

		if(t[rt].lazy) pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1;

		if(l <= mid) update(lson, l, r, val);

		if(r > mid) update(rson, l, r, val);

		pushup(rt); return;

	}

	int query(int rt, int l, int r) {

		if(l <= t[rt].l && t[rt].r <= r) { return t[rt].w; }

		if(t[rt].lazy) pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1, ans = 0;

		if(l <= mid) ans += query(lson, l, r);

		if(r > mid) ans += query(rson, l, r);

		return ans;

	}

}

signed main() {

	n = read(), m = read();

	Seg::build(1, 1, n);

	while(m--) {

		int opt = read(), x = read(), y = read(), k;

		if(opt == 1) k = read(), Seg::update(1, x, y, k);

		else cout << Seg::query(1, x, y) << '\n';

	}

	return 0;

}

区间加，区间乘，区间求和

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define int long long

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, mod;

namespace Seg {

	#define lson rt << 1

	#define rson rt << 1 | 1

	struct tre {

		int l, r, w, lazy1, lazy2;

	} t[A << 2];

	inline void pushup(int rt) {

		t[rt].w = (t[lson].w + t[rson].w) % mod;

	}

	inline void tag1(int rt, int val) {

		t[rt].lazy1 *= val, t[rt].lazy2 *= val;

		t[rt].lazy1 %= mod, t[rt].lazy2 %= mod;

		t[rt].w *= val, t[rt].w %= mod;

	}

	inline void tag2(int rt, int val) {

		t[rt].lazy2 += val; t[rt].lazy2 %= mod;

		t[rt].w += (t[rt].r - t[rt].l + 1) * val, t[rt].w %= mod;

	}

	inline void pushdown(int rt) {

		if(t[rt].lazy1 != 1) {

			tag1(lson, t[rt].lazy1);

			tag1(rson, t[rt].lazy1);

			t[rt].lazy1 = 1;

		}

		if(t[rt].lazy2) {

			tag2(lson, t[rt].lazy2);

			tag2(rson, t[rt].lazy2);

			t[rt].lazy2 = 0;

		}

	}

	inline void build(int rt, int l, int r) {

		t[rt].l = l, t[rt].r = r, t[rt].lazy1 = 1, t[rt].lazy2 = 0;

		if(l == r) { t[rt].w = read(); return; }

		int mid = (l + r) >> 1;

		build(lson, l, mid), build(rson, mid + 1, r);

		pushup(rt); return;

	}

	inline void mul(int rt, int l, int r, int val) {

		if(l <= t[rt].l && t[rt].r <= r) return tag1(rt, val);

		pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1;

		if(l <= mid) mul(lson, l, r, val);

		if(r > mid) mul(rson, l, r, val);

		pushup(rt); return;

	}

	inline void update(int rt, int l, int r, int val) {

		if(l <= t[rt].l && t[rt].r <= r) return tag2(rt, val);

		pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1;

		if(l <= mid) update(lson, l, r, val);

		if(r > mid) update(rson, l, r, val);

		pushup(rt); return;

	}

	inline int query(int rt, int l, int r) {

		if(l <= t[rt].l && t[rt].r <= r) { return t[rt].w % mod; }

		pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1, ans = 0;

		if(l <= mid) ans += query(lson, l, r), ans %= mod;

		if(r > mid) ans += query(rson, l, r), ans %= mod;

		return ans % mod;

	}

}

signed main() {

	n = read(), m = read(), mod = read();

	Seg::build(1, 1, n);

	while(m--) {

		int opt = read(), x = read(), y = read(), k;

		if(opt == 1) k = read() % mod, Seg::mul(1, x, y, k);

		else if(opt == 2) k = read() % mod, Seg::update(1, x, y, k);

		else cout << Seg::query(1, x, y) % mod << '\n';

	}

	return 0;

}

单调栈

for(int i = 0; i < T.size(); i++){

  while(! stk.empty() && stk.top() > T[i]){

    ​stk.pop();

  }

  stk.push(A[i]);

}

单调队列

上经典的滑动窗口问题

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#define maxn 1000100

using namespace std;

int q[maxn], a[maxn];

int n, k;

void getmin() {

	int head = 0, tail = 0;

	for (int i = 1; i < k; i++) {

		while (head <= tail && a[q[tail]] >= a[i]) tail--;

		q[++tail] = i;

	}

	for (int i = k; i <= n; i++) {

		while (head <= tail && a[q[tail]] >= a[i]) tail--;

		q[++tail] = i;

		while (q[head] <= i - k) head++;

		printf("%d ", a[q[head]]);

	}

}

void getmax() {

	int head = 0, tail = 0;

	for (int i = 1; i < k; i++) {

		while (head <= tail && a[q[tail]] <= a[i]) tail--;

		q[++tail] = i;

	}

	for (int i = k; i <= n; i++) {

		while (head <= tail && a[q[tail]] <= a[i]) tail--;

		q[++tail] = i;

		while (q[head] <= i - k) head++;

		printf("%d ", a[q[head]]);

	}

}

int main() {

	scanf("%d%d", &n, &k);

	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

	getmin();

	printf("\n");

	getmax();

	printf("\n");

	return 0;

}

树剖

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e6 + 11;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

int n, m, root, mod, w[A], pre[A];

struct node {

	int to, nxt;

} e[A];

int head[A], cnt;

inline void add_edge(int from, int to) {

	e[++cnt].to = to;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

namespace Seg {

	struct tree {

		int l, r, w, lazy;

	} t[A];

	#define lson rt << 1

	#define rson rt << 1 | 1

	inline void pushup(int rt) {

		t[rt].w = (t[lson].w + t[rson].w) % mod;

	}

	inline void pushdown(int rt) {

		t[lson].lazy += t[rt].lazy, t[rson].lazy += t[rt].lazy;

		t[lson].lazy %= mod, t[rson].lazy %= mod;

		t[lson].w += (t[lson].r - t[lson].l + 1) * t[rt].lazy; t[lson].w %= mod;

		t[rson].w += (t[rson].r - t[rson].l + 1) * t[rt].lazy; t[lson].w %= mod;

		t[rt].lazy = 0;

	}

	void build(int rt, int l, int r) {

		t[rt].l = l, t[rt].r = r;

		if(l == r) { t[rt].w = w[pre[l]] % mod; return; }

		int mid = (l + r) >> 1;

		build(lson, l, mid); build(rson, mid + 1, r);

		pushup(rt); return;

	}

	void add(int rt, int l, int r, int val) {

		if(l <= t[rt].l && t[rt].r <= r) {

			t[rt].lazy = (t[rt].lazy + val) % mod;

			t[rt].w = (t[rt].w + (t[rt].r - t[rt].l + 1) * val) % mod;

			return;

		}

		if(t[rt].lazy) pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1;

		if(l <= mid) add(lson, l, r, val);

		if(r > mid) add(rson, l, r, val);

		pushup(rt);

	}

	int asksum(int rt, int l, int r) {

		if(l <= t[rt].l && t[rt].r <= r) {

			return t[rt].w % mod;

		}

		if(t[rt].lazy) pushdown(rt);

		int mid = (t[rt].l + t[rt].r) >> 1, ans = 0;

		if(l <= mid) ans += asksum(lson, l, r);

		if(r > mid) ans += asksum(rson, l, r);

		return ans;

	}

}

int dfn[A], son[A], siz[A], dep[A], fa[A], top[A], tot;

void dfs1(int now, int fr) {

	siz[now] = 1, fa[now] = fr, dep[now] = dep[fr] + 1;

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(to == fr) continue;

		dfs1(to, now);

		siz[now] += siz[to];

		if(siz[to] > siz[son[now]]) son[now] = to;

	}

}

void dfs2(int now, int tp) {

	dfn[now] = ++tot, pre[tot] = now, top[now] = tp;

	if(son[now]) dfs2(son[now], tp);

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(to == fa[now] || to == son[now]) continue;

		dfs2(to, to);

	}

}

inline void add_qwq(int x, int y, int val) {

	while(top[x] != top[y]) {

		if(dep[top[x]] < dep[top[y]]) swap(x, y);

		Seg::add(1, dfn[top[x]], dfn[x], val);

		x = fa[top[x]];

	}

	if(dep[x] > dep[y]) swap(x, y);

	Seg::add(1, dfn[x], dfn[y], val);

	return;

}

inline int asksum_qwq(int x, int y) {

	int ans = 0;

	while(top[x] != top[y]) {

		if(dep[top[x]] < dep[top[y]]) swap(x, y);

		ans += Seg::asksum(1, dfn[top[x]], dfn[x]);

		ans %= mod;

		x = fa[top[x]];

	}

	if(dep[x] > dep[y]) swap(x, y);

	ans += Seg::asksum(1, dfn[x], dfn[y]);

	ans %= mod;

	return ans % mod;

}

int main() {

	n = read(), m = read(), root = read(), mod = read();

	for(int i = 1; i <= n; i++) w[i] = read() % mod;

	for(int i = 1; i < n; i++) {

		int x = read(), y = read();

		add_edge(x, y), add_edge(y, x);

	}

	dfs1(root, 0);

	dfs2(root, root);

	Seg::build(1, 1, n);

	while(m--) {

		int opt = read(), x, y, z;

		if(opt == 1) x = read(), y = read(), z = read() % mod, add_qwq(x, y, z);

		if(opt == 2) x = read(), y = read(), cout << asksum_qwq(x, y) % mod << '\n';

		if(opt == 3) x = read(), z = read(), Seg::add(1, dfn[x], dfn[x] + siz[x] - 1, z);

		if(opt == 4) x = read(), cout << Seg::asksum(1, dfn[x], dfn[x] + siz[x] - 1) % mod << '\n';

	}

	return 0;

}

求LCA

树剖求LCA

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 5e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, R;

struct node { int to, nxt; } e[A << 1];

int head[A], cnt, dep[A], son[A], dfn[A], siz[A], fa[A], top[A], tot;

inline void add(int from, int to) {

	e[++cnt].to = to;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

void prepare(int now, int fr) {

	fa[now] = fr, dep[now] = dep[fr] + 1, siz[now] = 1;

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(to == fr) continue;

		prepare(to, now); siz[now] += siz[to];

		if(siz[to] > siz[son[now]]) son[now] = to;

	}

}

void dfs(int now, int tp) {

	top[now] = tp, dfn[now] = ++tot;

	if(son[now]) dfs(son[now], tp);

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(to == fa[now] || to == son[now]) continue;

		dfs(to, to);

	}

}

inline int LCA(int x, int y) {

	while(top[x] != top[y]) {

		if(dep[top[x]] < dep[top[y]]) swap(x, y);

		x = fa[top[x]];

	}

	if(dep[x] > dep[y]) swap(x, y);

	return x;

}

int main() {

	n = read(), m = read(), R = read();

	for(int i = 1, x, y; i < n; i++) {

		x = read(), y = read();

		add(x, y), add(y, x);

	}

	prepare(R, 0); dfs(R, R);

	while(m--) {

		int x = read(), y = read();

		cout << LCA(x, y) << '\n';

	}

	return 0;

}

倍增求LCA

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 5e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, R;

struct node { int to, nxt; } e[A << 1];

int head[A], cnt, dep[A], fa[A][23], lg[A];

inline void add(int from, int to) {

	e[++cnt].to = to;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

void dfs(int now, int fr) {

	fa[now][0] = fr, dep[now] = dep[fr] + 1;

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(to == fr) continue;

		dfs(to, now);

	}

}

inline int LCA(int x, int y) {

	if(dep[x] < dep[y]) swap(x, y);

	for(int i = 20; i >= 0; i--) if(dep[fa[x][i]] >= dep[y]) x = fa[x][i];

	if(x == y) return x;

	for(int i = 20; i >= 0; i--) if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];

	return fa[x][0];

}

int main() {

	n = read(), m = read(), R = read(); lg[1] = 0;

	for(int i = 2, x, y; i <= n; i++) {

		x = read(), y = read();

		add(x, y), add(y, x);

		lg[i] = lg[i >> 1] + 1;

	}

	dfs(R, 0); 

	for(int j = 1; (1 << j) <= n; j++)

		for(int i = 1; i <= n; i++)

			fa[i][j] = fa[fa[i][j - 1]][j - 1];

	while(m--) {

		int x = read(), y = read();

		cout << LCA(x, y) << '\n';

	}

	return 0;

}

ST表

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, a[B], f[B][25], lg[B];

inline int query(int l, int r) {

	int k = lg[r - l + 1];

	return max(f[l][k], f[r - (1 << k) + 1][k]);

}

int main() {

	n = read(), m = read(); lg[0] = -1;

	for(int i = 1; i <= n; i++) a[i] = read(), f[i][0] = a[i], lg[i] = lg[i >> 1] + 1;

	for(int j = 1; (1 << j) <= n; j++)

		for(int i = 1; i <= n; i++)

			f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);

	while(m--) {

		int x = read(), y = read();

		cout << query(x, y) << '\n';

	}

	return 0;

}

数学/数论

求逆元

递推1~n的逆元

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define ll long long

using namespace std;

const int A = 1e5 + 11;

const int B = 3e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, p;

ll inv[B];

int main() {

	n = read(), p = read();

	inv[1] = 1;

	for(int i = 2; i <= n; i++) {

		inv[i] = 1LL * (p - p / i) * inv[p % i] % p;

	}

	for(int i = 1; i <= n; i++) cout << inv[i] << '\n';

	return 0;

}

矩阵快速幂

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define int long long

using namespace std;

const int A = 1e2 + 11;

const int B = 1e6 + 11;

const int mod = 1e9 + 7;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x =  + (c ^ 48);

	return x * f;

}

int n, k;

struct matrix {

	int a[A][A];

	matrix() { memset(a, 0, sizeof(a)); }

	void init() { for(int i = 1; i <= n; i++) a[i][i] = 1; }

} ans, a;

matrix operator * (const matrix &x, const matrix &y) {

	matrix qwq;

	for(int k = 1; k <= n; k++)

		for(int i = 1; i <= n; i++)

			for(int j = 1; j <= n; j++)

				qwq.a[i][j] = (qwq.a[i][j] + x.a[i][k] * y.a[k][j] % mod) % mod;

	return qwq;

}

inline void power() {

	while(k) {

		if(k & 1) ans = ans * a;

		a = a * a; k >>= 1;

	}

}

signed main() {

	n = read(), k = read();

	for(int i = 1; i <= n; i++)

		for(int j = 1; j <= n; j++)

			a.a[i][j] = read();

	ans.init();

	power();

	for(int i = 1; i <= n; i++) {

		for(int j = 1; j <= n; j++) {

			cout << ans.a[i][j] << " ";

		}

		puts("");

	}

	return 0;

}

快速乘

//知识点:快速乘

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

#define int long long

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

int a, b, p, ans;

int mul(int a, int b, int mod) {

	int res = 0;

	while(b) {

		if(b & 1) res = (res + a) % mod;

		a = (a + a) % mod;

		b >>= 1;

	}

	return res % mod;

}

signed main() {

	a = read(), b = read(), p = read();

	ans = mul(a, b, p);

	cout << ans << '\n';

	return 0;

}

快速幂

//知识点:快速幂

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

#define int long long

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

int a, b, p, ans;

int power(int a, int b, int mod) {

	int res = 1;

	while(b) {

		if(b & 1) res = res * a % mod;

		a = a * a % mod;

		b >>= 1;

	}

	return res % mod;

}

signed main() {

	a = read(), b = read(), p = read();

	ans = power(a, b, p);

	cout << ans << '\n';

	return 0;

}

埃氏筛

void prime(int n) {

	cnt = 0;

	vis[0] = vis[1] = 1;

	for(int i = 2; i <= n; ++i) {

		if(!vis[i]) {

			p[++cnt] = i;

			for(int j = i * i; j <= n; j+=i) {

				vis[j] = true;

			}

		}

	}

}

线性筛

int vis[N], p[N], cnt;

void prepare() {

	vis[0] = vis[1] = 1;

	for(int i = 2; i <= n; i++) {

		if(!vis[i]) p[++cnt] = i;

		for(int j = 1; j <= cnt; j++) {

			if(i * p[j] > n) break;

			vis[i * p[j]] = 1;

			if(i % p[j] == 0) break;

		}

	}

}

动态规划

最长上升子序列

普通二分

//知识点:

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 10001;

int f[N], a[N], n;

int len = 1;

int main() {

	memset(f, 0x3f, sizeof(f));

	n = read();

	for(int i = 1; i <= n; i++) a[i] = read();

	f[len] = a[1];

	for(int i = 2; i <= n; i++) {

		int l = 1, r = len, mid, rec;

		if(a[i] > f[len]) f[++len] = a[i];

		else {

			while(l <= r) {

				mid = (l + r) / 2;

				if(f[mid] >= a[i]) rec = mid, r = mid - 1;

				else l = mid + 1;

			}

			f[rec] = min(f[rec], a[i]);

		}

	}

	for(int i = 1; i <= len; i++) cout << f[i] << " ";

	cout << '\n' << len << '\n';

	return 0;

}

/*

9

7 2 1 5 6 4 3 8 9

*/

lower_bound

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 10001;

int f[N], a[N], n;

int len = 1;

int main() {

	memset(f, 0x3f, sizeof(f));

	n = read();

	for(int i = 1; i <= n; i++) a[i] = read();

	f[1] = a[1];

	for(int i = 2; i <= n; i++) {

		if(a[i] > f[len]) f[++len] = a[i];

		else {

			int p = lower_bound(f + 1, f + len + 1 , a[i]) - f;

			f[p] = a[i];

		}

	}

	for(int i = 1; i <= len; i++) cout << f[i] << " ";

	cout << '\n' << len << '\n';

	return 0;

}

最长公共子序列

\(n^2\)

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 1011;

int dp[N][N], a1[N], a2[N], n, m;

int main() {

	n = read();

	for(int i = 1; i <= n; i++) a1[i] = read();

	for(int i = 1; i <= n; i++) a2[i] = read();

	for(int i = 1; i <= n; i++)

		for(int j = 1; j <= n; j++) {

			dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

			if(a1[i] == a2[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);

		}

	cout << dp[n][n] << '\n';

	return 0;

}

图论

tarjan

#include <bits/stdc++.h>

using namespace std;

inline int read() {

    char c = getchar();

    int x = 0, f = 1;

    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

    for( ; isdigit(c); c = getchar()) x = x * 10 + c - 48;

    return x * f;

}

const int N = 1011;

struct node {

    int to, nxt;

} e[N];

int low[N], dfn[N], tot, topp = 0, sta[N], cnt, head[N];

bool vis[N];

inline void add(int from, int to) {

    e[++tot].to = to;

    e[tot].nxt = head[from];

    head[from] = tot;

}

void tarjan(int u) {

    dfn[u] = low[u] = ++cnt, sta[++topp] = u, vis[u] = 1;

    for(int i = head[u]; i; i = e[i].nxt) {

        int v = e[i].to;

        if(!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);

        else if(vis[v]) low[u] = min(low[u], dfn[v]);

    }

    if(dfn[u] == low[u]) {

        do {

            cout << sta[topp] << ' ';

            vis[sta[topp--]] = 0;

        }while(u != sta[topp + 1]);

        cout << '\n';

    }

}

int n, m;

int main() {

    n = read(), m = read();

    for(int i = 1; i <= m; i++) {

        int u = read(), v = read();

        add(u, v);

    }

    for(int i = 1; i <= n; i++)

        if(!dfn[i]) tarjan(i);

}

/*

5 8

1 2

2 3

3 6

5 6

1 4

4 5

5 1

2 5

*/

最小生成树

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 2e5 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

int n, m, fa[A], cnt, ans;

struct node { int x, y, val; } e[A];

int find(int x){ return fa[x] == x ? x : fa[x] = find(fa[x]); }

bool cmp(node x, node y) { return x.val < y.val; }

int main() {

	n = read(), m = read();

	for(int i = 1; i <= n; i++) fa[i] = i;

	for(int i = 1; i <= m; i++) e[i].x = read(), e[i].y = read(), e[i].val = read();

	sort(e + 1, e + 1 + m, cmp);

	for(int i = 1; i <= m; i++) {

		int dx = find(e[i].x), dy = find(e[i].y);

		if(dx != dy) {

			fa[dy] = dx;

			ans += e[i].val;

			cnt++;

			if(cnt == n - 1) break;

		}

	}

	cout << ans << '\n';

	return 0;

}

求树的直径

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

inline int read() {

    char c = getchar();

    int x = 0, f = 1;

    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

    return x * f;

}

const int M = 3e5 + 11;

int n, cnt, k, ans, head[M], dep[M], vis[M];

struct node {

    int nxt, to;

} edge[M];

void add(int from, int to) {

    edge[++cnt].nxt = head[from];

    edge[cnt].to = to;

    head[from] = cnt;

}

void dfs(int x, int fx) {

    vis[x] = 1;

    for(int i = head[x]; i; i = edge[i].nxt) {

        int to = edge[i].to;

        if(to == fx) continue;

        dep[to] = dep[x] + 1;

        if(ans < dep[to]) {

            ans = dep[to];

            k = to;

        }

        dfs(to, x);

    }

}

int main() {

    n = read();

    for(int i = 1; i < n; i++) {

        int x = read(), y = read();

        add(x, y), add(y, x);

        ans = 0, dep[x] = 0, memset(vis, 0, sizeof(vis)); dfs(x, 0);

        ans = 0, dep[k] = 0, memset(vis, 0, sizeof(vis)); dfs(k, 0);

        cout << ans << '\n';

    }

    return 0;

}

二分图匹配

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e6 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, E, ans;

struct node { int to, nxt; } e[A];

int head[A], cnt, match[A], vis[A];

inline void add(int from, int to) {

	e[++cnt].to = to;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

bool dfs(int now) {

	for(int i = head[now]; i; i = e[i].nxt) {

		int to = e[i].to;

		if(vis[to]) continue;

		vis[to] = 1;

		if(!match[to] || dfs(match[to])) {

			match[to] = now;

			return 1;

		}

	}

	return 0;

} 

int main() {

	n = read(), m = read(), E = read();

	for(int i = 1, x, y; i <= E; i++) {

		x = read(), y = read();

		if(x > n || y > m) continue;

		add(x, y);

	}

	for(int i = 1; i <= n; i++) {

		memset(vis, 0, sizeof(vis));

		if(dfs(i)) ans++;

	}

	cout << ans << '\n';

	return 0;

}

最短路

堆优化dij

#include <queue>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

const int inf = 0x3f3f3f3f;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, dis[A], S;

struct node { int to, nxt, val; } e[B];

int head[A], cnt;

inline void add(int from, int to, int val) {

	e[++cnt].to = to;

	e[cnt].val = val;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

struct edge {

	int x, y;

	bool operator < (const edge &qwq) const {

		return y > qwq.y;

	}

};

priority_queue<edge> Q;

inline void ZDL() {

	memset(dis, inf, sizeof(dis));

	dis[S] = 0;

	Q.push((edge) {S, 0});

	while(!Q.empty()) {

		int u = Q.top().x, d = Q.top().y; Q.pop();

		if(d != dis[u]) continue;

		for(int i = head[u]; i; i = e[i].nxt) {

			int to = e[i].to;

			if(dis[to] > dis[u] + e[i].val) {

				dis[to] = dis[u] + e[i].val;

				Q.push((edge) {to, dis[to]});

			}

		}

	}

}

int main() {

//	freopen("testdata.in", "r", stdin);

	n = read(), m = read(), S = read();

	for(int i = 1, x, y, val; i <= m; i++) {

		x = read(), y = read(), val = read();

		add(x, y, val);

	}

	ZDL();

	for(int i = 1; i <= n; i++) {

		if(dis[i] == inf) cout << "2147483647 ";

		else cout << dis[i] << " ";

	}

	return 0;

}

SPFA

#include <queue>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

const int inf = 0x3f3f3f3f;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, dis[A], S, vis[A];

struct node { int to, nxt, val; } e[B];

int head[A], cnt;

inline void add(int from, int to, int val) {

	e[++cnt].to = to;

	e[cnt].val = val;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

queue <int> Q;

inline void SPFA() {

	memset(dis, inf, sizeof(dis));

	dis[S] = 0, vis[S] = 1;

	Q.push(S);

	while(!Q.empty()) {

		int u = Q.front(); Q.pop(); vis[u] = 0;

		for(int i = head[u]; i; i = e[i].nxt) {

			int to = e[i].to;

			if(dis[to] > dis[u] + e[i].val) {

				dis[to] = dis[u] + e[i].val;

				if(!vis[to]) {

					vis[to] = 1;

					Q.push(to);

				}

			}

		}

	}

}

signed main() {

	n = read(), m = read(), S = read();

	for(int i = 1, x, y, val; i <= m; i++) {

		x = read(), y = read(), val = read();

		add(x, y, val);

	}

	SPFA();

	for(int i = 1; i <= n; i++) cout << dis[i] << " ";

	return 0;

}

判负环

#include <queue>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e5 + 11;

const int B = 1e6 + 11;

const int inf = 0x3f3f3f3f;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, dis[A], S = 1, vis[A], tot[A];

struct node { int to, nxt, val; } e[B];

int head[A], cnt;

inline void add(int from, int to, int val) {

	e[++cnt].to = to;

	e[cnt].val = val;

	e[cnt].nxt = head[from];

	head[from] = cnt;

}

inline bool SPFA() {

	memset(dis, inf, sizeof(dis));

	dis[S] = 0, vis[S] = 1, tot[S] = 1;

	queue <int> Q;

	Q.push(S);

	while(!Q.empty()) {

		int u = Q.front(); Q.pop(); vis[u] = 0;

		for(int i = head[u]; i; i = e[i].nxt) {

			int to = e[i].to;

			if(dis[to] > dis[u] + e[i].val) {

				if(++tot[to] >= n) return 1;

				dis[to] = dis[u] + e[i].val;

				if(!vis[to]) {

					vis[to] = 1;

					Q.push(to);

				}

			}

		}

	}

	return 0;

}

signed main() {

	int T = read();

	while(T--) {

		memset(head, 0, sizeof(head));

		memset(vis, 0, sizeof(vis));

		memset(tot, 0, sizeof(tot));

		n = read(), m = read();

		for(int i = 1, x, y, val; i <= m; i++) {

			x = read(), y = read(), val = read();

			add(x, y, val);

			if(val >= 0) add(y, x, val);

		}

		if(SPFA()) puts("YE5");

		else puts("N0");

	}

	return 0;

}

字符串

\(hash\)

/*

By:Loceaner

*/

#include <cstdio>

#include <cstring>

#include <iostream>

#define ull unsigned long long

using namespace std;

inline int read() {

	char c = getchar();

	int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);

	return x * f;

}

const int N = 211;

char pa[N][N], meng[N];

int n, m, a[N], l[N], ans, cnt[N];

ull p[N], cnm[N][N], sjp[N];

bool query(int l1, int r1,int now) {

	ull h1, h2;

	h1 = sjp[r1] - sjp[l1 - 1] * p[r1 - l1 + 1];

	h2 = cnm[now][l[now]];

	return h1 == h2;

}

void work2() {//cnm

	memset(sjp, 0, sizeof(sjp));

	int len = strlen(meng + 1);

	for(int i = 1; i <= len; i++) {

		sjp[i] = sjp[i - 1] * 27 + meng[i] - 'a' + 1;

	}

	for(int i = 1; i <= m; i++) {

		for(int j = 1; j <= len; j++) {

			if(meng[j] == pa[i][1]) {

				if(query(j, j + l[i] - 1, i)) ans += j * a[i];

			}

		}

	}

}

int main() {

	freopen("dream.in", "r", stdin);

	freopen("dream.out", "w", stdout);

	n = read(), m = read();

	p[0] = 1;

	for(int i = 1; i <= 200; i++) p[i] = p[i - 1] * 27ull;

	for(int i = 1; i <= m; i++) scanf("%s", pa[i] + 1), l[i] = strlen(pa[i] + 1);

	for(int i = 1; i <= m; i++)

		for(int j = 1; j <= l[i]; j++)

			cnm[i][j] = cnm[i][j - 1] * 27 + pa[i][j] - 'a' + 1;

	for(int i = 1; i <= m; i++) a[i] = read();

	for(int i = 1; i <= n; i++) scanf("%s", meng + 1), work2();

	cout << ans << '\n';

	return 0;

}

KMP

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int A = 1e6 + 11;

const int B = 1e6 + 11;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);

	return x * f;

}

int n, m, nxt[A];

char s[A], t[A];

void get_nxt() {

	int p = 0;

	for(int i = 2; i <= m; i++) {

		while(p && t[i] != t[p + 1]) p = nxt[p];

		if(t[i] == t[p + 1]) ++p;

		nxt[i] = p;

	}

}

int main() {

	scanf("%s", s + 1); scanf("%s", t + 1);

	n = strlen(s + 1), m = strlen(t + 1);

	get_nxt();

	int p = 0;

	for(int i = 1; i <= n; i++) {

		while(p && s[i] != t[p + 1]) p = nxt[p];

		if(s[i] == t[p + 1]) {

			++p;

			if(p == m) {

				cout << i - m + 1 << '\n';

				p = nxt[p];

			}

		}

	}

	for(int i = 1; i <= m; i++) cout << nxt[i] << " ";

	return 0;

}

CSP考前复习的更多相关文章

  1. 考前复习(codevs 2837)

    2837 考前复习  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 黄金 Gold 题解  查看运行结果     题目描述 Description Aiden马上要考试了,可他 ...

  2. CSP 初赛复习 密码

    CSP 初赛复习 密码是\(xj\)机房学生端密码

  3. CSP初赛复习

    初赛复习 初赛一定要过啊,否则付出的那么多都白搭了! while(1) ++csp.rp,++csp.luck,++csp.scores; 历史 2020年开始,除NOIP以外的NOI系列其他赛事(包 ...

  4. Codevs 2837 考前复习

     时间限制: 1 s  空间限制: 128000 KB  题目等级 : 黄金 Gold 题解  查看运行结果     题目描述 Description Aiden马上要考试了,可他还没怎么复习,于是他 ...

  5. codevs 2837 考前复习——01背包

     时间限制: 1 s  空间限制: 128000 KB  题目等级 : 黄金 Gold 题解  查看运行结果     题目描述 Description Aiden马上要考试了,可他还没怎么复习,于是他 ...

  6. NOIP&CSP 考前 Dev-cpp 的选项问题和考试心态

    (进入考场后您将获得一个崭新的 \(Dev-cpp\),没有中文,没有编译选项,没有缺省源:我还将获得一个崭新的脑子,没有心态,没有智商,没有调试能力--) 中文 \[Step1 \] \[Step2 ...

  7. NOIP考前复习-数制转换，数论模板与文件读写

    数制转换有两种题型,一般一题,分值1.5分. 题型一:R进制转十进制 解法就是:按权展开,但要注意各个位的权,最低位(最右边)的权是0次方,权值为1. 纯整数的情况: (11010110)2 = 1× ...

  8. CSP2019 考前复习

    动态规划 [NOIP2016]愤怒的小鸟(状压+思维) 多组数据题 共有i只猪,给出每只猪的坐标,鸟的飞行轨迹为经过原点的抛物线,求最少要多少只鸟能消灭所有的猪 \[ 猪数量n<=18 \] 看 ...

  9. NOIP的模板--考前复习

    距离NOIP还有25天 可以去放弃一些巨难得题目去搞一些模板了 -------在校老师的原话 一·快排 虽然可以手打,最好用STL,里面有很多优化,会快很多 #include<iostream& ...

随机推荐

  1. Identity Server4资料

    https://www.cnblogs.com/cgzl/p/9405796.html https://www.cnblogs.com/cgzl/p/7780559.html https://clou ...

  2. [开源]OSharpNS 步步为营系列 - 4. 添加业务对外API

    什么是OSharp OSharpNS全称OSharp Framework with .NetStandard2.0,是一个基于.NetStandard2.0开发的一个.NetCore快速开发框架.这个 ...

  3. HTML连载40-盒子宽度和高度的练习、box-sizing属性

    一.判断方法 1.判断是否元素宽高为200的盒子 只需要看:边框+内边距+内容宽度/内容高度的值是否等于200 2.判断是否内容宽高为100的盒子 只需要看:width和heght的值是否等于100 ...

  4. WPF-数据模板深入(加载XML类型数据)

    一.我们知道WPF数据模板是当我们给定一个数据类型,我们为这个数据类型写好布局,就给这种数据类型穿上了外衣. 下面这个例子,能够帮助大家充分理解数据模板就是数据类型的外衣的意思:(里面的MyListB ...

  5. DevExpress的TreeList怎样设置数据源，从实例入手

    场景 Winform控件-DevExpress18下载安装注册以及在VS中使用: https://blog.csdn.net/BADAO_LIUMANG_QIZHI/article/details/1 ...

  6. WPF 基础总结(学习建议)

    举个简单得例子, 类似造房子, 当然实际上可能非常细, 对应的如下所示: 在此之前, 需要了解的是. WPF项目是怎么启动的 Xaml的结构是怎么样组成, 命名控件定义引用的方法. 知道了如何在Xam ...

  7. 一款对Postman支持较好的接口文档生成工具

    最近要编写接口文档给测试和前端看,通过网上查阅资料,也认识了很多款接口文档生成工具,比如易文档.ApiPost.ShowDoc.YApi.EoLinker.DOClever.apizza等,通过对这几 ...

  8. Python 电子邮件

    从一台计算机编写邮件到对方收到邮件.假设我们自己的电子邮件地址是me@163.com,对方的电子邮件地址是friend@sina.com 我们在本地的软件上写好邮件,点击发送,邮件就发送出去了,这些电 ...

  9. Java面向对象——类的成员

    Java面向对象——类的成员 摘要:本文主要介绍了类的常见成员. 属性 属性称为成员变量,一般来讲不用赋值,因为有默认值,另外显式赋值没有意义会导致所有由此类创建对象都是此值. 默认值 Boolean ...

  10. Java学习——泛型

    Java学习——泛型 摘要:本文主要介绍了什么是泛型,为什么要用泛型,以及如何使用泛型. 部分内容来自以下博客: https://www.cnblogs.com/lwbqqyumidi/p/38376 ...