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「赛后总结」AtCoder Beginner Contest 177

作者：互联网

题意 & 题解

A Don't be late

题意：给你路程，时间限制，速度，问你能不能在时间限制内走路程那么远。
题解：数学题？物理题？

B Substring

题意：给你两个字符串 \(s,t\)，问你最少更改 \(s\) 几次能使得 \(t\) 是 \(s\) 的字串。
题解：因为字符串长度小于 \(1000\) 直接暴力枚举 \(t\) 出现在 \(s\) 的什么位置。

C Sum of product of pairs

题意：求 \(\sum\limits_{i = 1}^{N - 1}\sum\limits_{j = i + 1} ^ {N} A_i A_j\)
题解：数学题，化简一下就成了 \(\sum\limits_{i = 1}^{N - 1}A_i\sum\limits_{j = i + 1} ^ {N} A_j\)，后面用前缀和维护一下就好。

D Friends

题意：给你一些人之间的朋友关系，可以互相传递，问你最少分成多少组使得每一组都没有两个人是朋友。
题解：并查集。最大的连通块的大小就是答案，应该挺显然的。

E Coprime

题意：咕咕咕
题解：咕咕咕

F I hate Shortest Path Problem

题意：咕咕咕
题解：咕咕咕

Code

A Don't be late

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>

int a, b, c;

int main() {
	std::cin >> a >> b >> c;
	double t = (a * 1.0) / (c * 1.0);
	if (t - b * 1.0 >= 1e-8) puts("No");
	else puts("Yes");
	return 0;
}

B Substring

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>

int min(int a, int b) { return a < b ? a : b; }

int ans = 2147483647;
std::string s, t;

int main() {
	std::cin >> s >> t;
	int len1 = s.length(), len2 = t.length();
	for (int i = 0; i < len1; ++i) {
		if (i + len2 - 1 >= len1) break;
		int p = i, sum = 0;
		for (int j = 0; j < len2; ++j, ++p) {
			if (s[p] == t[j]) ++sum;
		}
		ans = min(ans, len2 - sum);
	}
	std::cout << ans << '\n';
	return 0;
}

C Sum of product of pairs

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define MAXN 200001
#define int long long

const int mod = 1000000007;
int n, ans, a[MAXN], sum[MAXN];

signed main() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
		sum[i] = (1ll * sum[i - 1] + 1ll * a[i]) % mod;
	}
	for (int i = 1; i < n; ++i) {
		ans = (ans + 1ll * a[i] * ((sum[n] - sum[i] + mod) % mod) % mod) % mod;
	}
	std::cout << ans << '\n';
	return 0;
}

D Friends

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define MAXN 200001

int max(int a, int b) { return a > b ? a : b; }

int n, m, fa[MAXN], size[MAXN];

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

void Union(int x, int y) {
	int rootx = find(x), rooty = find(y);
	if (rootx == rooty) return;
	if (size[rootx] > size[rooty]) {
		fa[rooty] = rootx;
		size[rootx] += size[rooty];
	}
	else {
		fa[rootx] = rooty;
		size[rooty] += size[rootx];
	}
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; ++i) fa[i] = i, size[i] = 1;
	for (int i = 1, u, v; i <= m; ++i) {
		scanf("%d %d", &u, &v);
		Union(u, v);
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		if (fa[i] == i) ans = max(ans, size[i]);
	}
	std::cout << ans << '\n';
	return 0;
}

rating & 总结

标签：AtCoder,题意,Beginner,Contest,int,题解,sum,rooty,include
来源： https://www.cnblogs.com/poi-bolg-poi/p/13583684.html