NOIP做题练习（day1）

A - Xenny and Alternating Tasks

题面

题解

枚举第一天是谁做，将两个答案取\(min\)即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
int t, n, a[20003], b[20003];
int main()
{
	//File("XENTASK");
	t = gi();
	while (t--)
	{
		n = gi();
		for (int i = 1; i <= n; i+=1) a[i] = gi();
		for (itn i = 1; i <= n; i+=1) b[i] = gi();
		itn ans = 0, sum = 0;
		for (itn j = 1; j <= n; j+=1)
		{
			if (j & 1) sum = sum + a[j];
			else sum = sum + b[j];
		}
		for (int j = 1; j <= n; j+=1)
		{
			if (j & 1) ans = ans + b[j];
			else ans = ans + a[j];
		}
		printf("%d\n", min(ans, sum));
	}
	return 0;
}

B - Bear and Extra Number

题面

题解

将数列排序，遍历数组元素，然后分类讨论：

1. 如果是\(a_i = a_{i+1}\) ，那么输出\(a_i\)。
2. 如果\(a_{i+1}-a_i>1\) ：
   • 因为题目说有唯一解，因此\(i\)只能为\(n-1\)或\(1\)。
     • 若\(i\)为\(n-1\)，则输出\(a_n\)。
     • 若\(i\)为\(1\)，则输出\(a_1\)。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
int t, n, a[100003], ans, sum;
int main()
{
	//File("EXTRAN");
	t = gi();
	while (t--)
	{
		n = gi();
		for (itn i = 1; i <= n; i+=1) a[i] = gi();
		sort(a + 1, a + 1 + n);
		for (int i = 1; i < n; i+=1)
		{
			if (a[i + 1] == a[i]) {printf("%d\n", a[i]); break;}
			if (a[i + 1] - a[i] > 1)
			{
				if (i == n - 1) {printf("%d\n", a[n]); break;}
				else if (i == 1) {printf("%d\n", a[1]); break;}
			}
		}
	}
	return 0;
}

C - Cooking Schedule

题面

题解

二分。

注意\(check\)怎么写。

第一次做时没想到二分，很遗憾。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
int n, m, t, k, ans, kkk, c[1000003];
char s[1000003];
int CCC(itn x)
{
	int u = 0;
	for (int i = 1; i <= n; i+=1)
	{
		if (i % 2 == x)
		{
			if (s[i] == '0') ++u;
		}
		else {if (s[i] == '1') ++u;}
	}
	return u;
}
bool check(int x)
{
	if (x != 1)
	{
		int u = 0;
		for (itn i = 1; i <= kkk; i+=1) if (c[i] > x) u = u + c[i] / (x + 1);
		return u <= k;
	}
	else {int u = min(CCC(1), CCC(0)); return u <= k;}
}
signed main()
{
	//File("SCHEDULE");
	t = gi();
	while (t--)
	{
		int l = 1, r = 0, uuu = 0; kkk = 0;
		n = gi(), k = gi();
		scanf("%s", s + 1);
		for (int i = 1; i <= n; i+=1)
		{
			if (s[i] != s[i + 1])
			{
				c[++kkk] = i - uuu; uuu = i; r = max(r, c[kkk]);
			}
		}
		while (l < r)
		{
			int mid = (l + r) >> 1;
			if (check(mid)) r = mid;
			else l = mid + 1;
		}
		printf("%lld\n", r);
	}
	return 0;
}

D - Subtree Removal

题面

题解

简单树形\(DP\)。

看似很难，实际代码很短。

计算出每棵子树的大小，与\(-X\)取\(max\)即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
int t, n, x, fa[100003], ans, sum, k, b[100003], sz[100003];
vector <int> e[100003];
int dfs(int u, int fa)
{
	int o = b[u];
	for (int i = e[u].size() - 1; ~i; i-=1)
	{
		int v = e[u][i];
		if (v == fa) continue;
		o = o + dfs(v, u);
	}
	return max(o, -x);
}	   
signed main()
{
	//File("SUBREM");
	t = gi();
	while (t--)
	{
		n = gi(), x = gi();
		int sum = 0;
		for (itn i = 1; i <= n; i+=1) b[i] = gi(), sz[i] = b[i], sum = sum + b[i], e[i].clear();
		for (itn i = 1; i < n; i+=1) {int u = gi(), v = gi(); e[u].push_back(v), e[v].push_back(u);}
		printf("%lld\n", dfs(1, -1));
	}
	return 0;
}

E - Dish Owner

题面

题解

并查集简单题。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#include <queue>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
itn t, n, s[10003], fl, x, y, Q;
itn a[10003], fa[10003];
int getf(itn u) {if (u == fa[u]) return u; return fa[u] = getf(fa[u]);}
inline void unionn(int u, int v) {
	if (a[u] > a[v])
	{
		fa[v] = u;
	}
	else if (a[v] > a[u])
	{
		fa[u] = v;
	}
}
int main()
{
	//File("DISHOWN");
	t = gi();
	while (t--)
	{
		n = gi();
		for (itn i = 1; i <= n; i+=1) s[i] = gi(), a[i] = s[i], fa[i] = i;
		Q = gi();
		for (itn i = 1; i <= Q; i+=1)
		{
			fl = gi();
			if (fl)
			{
				x = gi();
				printf("%d\n", getf(x));
			}
			else
			{
				x = gi(), y = gi();
				int X = getf(x), Y = getf(y);
				if (X == Y) puts("Invalid query!");
				else unionn(X, Y);
			}
		}
	}
	return 0;
}

F - Triplets

题面

题解

数学题。

大暴力很好写，考虑如何优化。

预处理出\(a\)、\(c\)序列的前缀和。

逐一枚举\(b\)序列上的数，找出\(a\)、\(c\)序列中比当前枚举到的数小的数。

计算出其对答案的贡献即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}
const int mod = 1000000007;
itn t, n, p, q, r, a[100003], b[100003], c[100003], x, y, z;
itn s1 = 1, s2 = 1, s3 = 1;
signed main()
{
	//File("SUMQ");
	t = gi();
	while (t--)
	{
		p = gi(), q = gi(), r = gi();
		for (int i = 1; i <= p; i+=1) a[i] = gi();
		for (int i = 1; i <= q; i+=1) b[i] = gi();
		for (itn i = 1; i <= r; i+=1) c[i] = gi();
		sort(a + 1, a + 1 + p); sort(b + 1, b + 1 + q); sort(c + 1, c + 1 + r);
		int p1 = 1, p2 = 1, c1 = 0, c2 = 0, s1 = 0, s2 = 0, ans = 0;
	   	for (int i = 1; i <= q; i+=1)
	   	{
		   	while (p1 <= p && a[p1] <= b[i])
		   	{
		   		s1 = (s1 + a[p1]) % mod;
		   		++c1, ++p1;
		   	}
			while (p2 <= r && c[p2] <= b[i])
		   	{
				s2 = (s2 + c[p2]) % mod;
			   	++c2, ++p2;
	   		}
	   		ans = (ans % mod + c1 * c2 % mod * b[i] % mod * b[i] % mod) % mod;
	   		ans = (ans % mod + b[i] * (s1 * c2 % mod + s2 * c1 % mod) % mod) % mod;
	   		ans = (ans % mod + s1 * s2 % mod) % mod;
	   	}
	   	printf("%lld\n", ans % mod);
	}
	return 0;
}

总结

这次练习第一次做的时候做得一般般。

做题的策略要更好一点。

细节地方要注意。