【AtCoder】KEYENCE Programming Contest 2019

A - Beginning

这个年份恐怕需要+2

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 80005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int num[10];
void Solve() {
	for(int i = 1 ; i <= 4 ; ++i) read(num[i]);
	sort(num + 1,num + 5);
	if(num[1] == 1 && num[2] == 4 && num[3] == 7 && num[4] == 9) {puts("YES");}
	else puts("NO");
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

B - KEYENCE String

……

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 80005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
string s;
string tar = "keyence";
void Solve() {
	cin >> s;
	int p = 0;
	while(s[p] == tar[p]) ++p;
	int q = 0;
	while(s[s.length() - 1 - q] == tar[tar.length() - 1 - q]) ++q;
	if(p + q >= tar.length()) puts("YES");
	else puts("NO");
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

C - Exam and Wizard

就是先用一遍A的总和恢复出B，如果不足就是-1，然后贪心每次恢复最少可以回到原来的大小

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;
int64 A[MAXN],B[MAXN],all,C[MAXN];
multiset<int64> S;
void Solve() {
	read(N);
	for(int i = 1 ; i <= N ; ++i) {
		read(A[i]);all += A[i];
	}
	for(int i = 1 ; i <= N ; ++i) {
		read(B[i]);all -= B[i];
		if(A[i] - B[i] >= 0) S.insert(A[i] - B[i]);
	}
	if(all < 0) {puts("-1");return;}
	int cnt = 0;
	while(!S.empty()) {
		int64 p = *S.begin();
		S.erase(S.begin());
		if(all >= p) {all -= p;++cnt;}
		else break;
	}
	out(N - cnt);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

D - Double Landscape

感觉这题so interesting 啊，还可以往高维拓展

从大到小往里面塞数，记录一下每次能用的行和列，如果这个数必须在某一行或某一列，两个都固定就乘上1，一个固定，比如行固定，就乘上能用的列数

如果这个数位置没被固定，就是乘上当前能用的行列数的乘积，减去比它大的数

为什么呢。。因为一个数能填的位置，要么和前面比它大的数都没有交集，要么前面比它大的数能填的集合是它的一个真子集

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 1005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M;
bool col[MAXN * MAXN],row[MAXN * MAXN];
int inc(int a,int b) {
	return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
	return 1LL * a * b % MOD;
}
void Solve() {
	read(N);read(M);
	int a;
	for(int i = 1 ; i <= N ; ++i) {
		read(a);
		if(row[a]) {puts("0");return;}
		row[a] = 1;
	}
	for(int i = 1 ; i <= M ; ++i) {
		read(a);
		if(col[a]) {puts("0");return;}
		col[a] = 1;
	}
	int ans = 1;
	int c = 0,r = 0;
	for(int i = N * M ; i >= 1 ; --i) {
		int tc = c,tr = r;
		if(col[i]) {tc = 1;++c;}
		if(row[i]) {tr = 1;++r;}
		if(col[i] || row[i]) ans = mul(ans,tc * tr);
		else ans = mul(ans,tc * tr - (N * M - i));
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

E - Connecting Cities

外国人真是瞧不起中国人的数据结构水平

随手写了个线段树就过了

就是拆式子，分\(-i * D + A_i\)和\(i * D + A_i\)

然后跑prim，两个线段树取最小的就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N;int64 D;
int64 A[MAXN];
struct seg_tr {
	struct node {
		int L,R;
		int64 cov;
		pair<int64,int> val[2];
	}tr[MAXN * 4];
	void update(int u) {
		tr[u].val[1] = mp(1e16,0);
		if(tr[u].L != tr[u].R) {
			tr[u].val[0] = min(tr[u << 1].val[0],tr[u << 1 | 1].val[0]);
			tr[u].val[1] = min(tr[u << 1].val[1],tr[u << 1 | 1].val[1]);
		}
		tr[u].val[1] = min(tr[u].val[1],mp(tr[u].cov + tr[u].val[0].fi,tr[u].val[0].se));
	}
	void build(int u,int l,int r,int on) {
		tr[u].L = l;tr[u].R = r;tr[u].cov = 1e16;
		if(l == r) {
			tr[u].val[0] = mp(A[l] + on * l * D,l);
			update(u);
			return;
		}
		int mid = (l + r) >> 1;
		build(u << 1,l,mid,on);build(u << 1 | 1,mid + 1,r,on);
		update(u);
	}
	void Change(int u,int l,int r,int64 v) {
		if(l > r) return;
		if(tr[u].L == l && tr[u].R == r) {
			tr[u].cov = min(tr[u].cov,v);
			update(u);
			return;
		}
		int mid = (tr[u].L + tr[u].R) >> 1;
		if(r <= mid) Change(u << 1,l,r,v);
		else if(l > mid) Change(u << 1 | 1,l,r,v);
		else {Change(u << 1,l,mid,v);Change(u << 1 | 1,mid + 1,r,v);}
		update(u);
	}
	void Change_pos(int u,int p) {
		if(tr[u].L == tr[u].R) {
			tr[u].val[0] = mp(1e16,tr[u].L);
			update(u);return;
		}
		int mid = (tr[u].L + tr[u].R) >> 1;
		if(p <= mid) Change_pos(u << 1,p);
		else Change_pos(u << 1 | 1,p);
		update(u);
	}
}s[2];
void Solve() {
	read(N);read(D);
	for(int i = 1 ; i <= N ; ++i) read(A[i]);
	s[0].build(1,1,N,1);
	s[1].build(1,1,N,-1);
	s[0].Change_pos(1,1);s[1].Change_pos(1,1);
	s[0].Change(1,2,N,A[1] - D);
	int cnt = N - 1;
	int64 ans = 0;
	while(cnt--) {
		pair<int64,int> t[2];
		t[0] = s[0].tr[1].val[1],t[1] = s[1].tr[1].val[1];
		if(t[0] > t[1]) swap(t[0],t[1]);
		ans += t[0].fi;
		int u = t[0].se;
		s[0].Change_pos(1,u);s[1].Change_pos(1,u);
		s[1].Change(1,1,u - 1,1LL * u * D + A[u]);
		s[0].Change(1,u + 1,N,-1LL * u * D + A[u]);
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}

F - Paper Cutting

怎么又是计数转期望……输了……水平不行，从前遇见期望只会转计数。。。直到atc用一堆计数转期望把我吊着打

就是考虑一个左下角为\((i,j)\)的方形，当这个点作为左下角后，每切一刀都会被统计一次，为了方便我们姑且只讨论\(i > 0\)且\(j > 0\)的情况

这个矩形第一次存在的概率是

\(\frac{\binom{K}{2}}{\binom{N}{2}}\)

就是考虑选\(i,j\)(无序）我们需要它是\(\binom{N}{2}\)中\(\binom{K}{2}\)对中的一个

之后被再次算的概率是，就是，我又拿了一个，左下角为\(i,j\)的就又被算了一次

\(\frac{\binom{K}{3}}{\binom{N}{3}}\cdot \frac{1}{3}\)我需要最后一个选的点在最后，这样的概率有\(\frac{1}{3}\)

这样的点得集合有\(N - 2\)种，所以最后的期望是

\(\frac{\binom{K}{2}}{\binom{N}{2}} + \frac{\binom{K}{3}}{\binom{N}{3}} \cdot \frac{1}{3} \cdot (N - 2)\)

对于边上的点和角上的一个点也是同理的

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
const int MOD = 1000000007;
int H,W,K,N;
int C[2][4],inv[4];
int inc(int a,int b) {
	return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
	return 1LL * a * b % MOD;
}
void update(int &x,int y) {
	x = inc(x,y);
}
int fpow(int x,int c) {
	int res = 1,t = x;
	while(c) {
		if(c & 1) res = mul(res,t);
		t = mul(t,t);
		c >>= 1;
	}
	return res;
}
void Solve() {
	read(H);read(W);read(K);
	N = H + W;
	inv[1] = 1;inv[2] = (MOD + 1) / 2;inv[3] = (MOD + 1) / 3;
	C[0][0] = C[1][0] = 1;
	for(int i = 1 ; i <= 3 ; ++i) {
		C[0][i] = mul(mul(C[0][i - 1],inv[i]),K - i + 1);
		C[1][i] = mul(mul(C[1][i - 1],inv[i]),N - i + 1);
	}
	int ans = K;
	int t1 = mul(K,fpow(N,MOD - 2)),t2 = mul(C[0][2],fpow(C[1][2],MOD - 2)),t3 = mul(C[0][3],fpow(C[1][3],MOD - 2));
	int s = inc(t1,mul(mul(t2,inv[2]),N - 1));
	update(ans,mul(s,H + W));
	s = inc(t2,mul(mul(t3,inv[3]),N - 2));
	update(ans,mul(s,1LL * H * W % MOD));
	for(int i = N ; i >= N - K + 1; --i) {
		ans = mul(ans,i);
	}
	out(ans);enter;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}