Educational Codeforces Round 31

最新推荐文章于 2018-01-28 11:57:49 发布

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本文提供了四道编程题目的解答思路及代码实现：第一题计算完成任务所需的最小天数；第二题判断序列是否存在特定组合；第三题求解图中最大可达点对；第四题最小化数组元素之和。

#include<bits/stdc++.h>

using namespace std;
#define LL long long
#define mst(a, b)	memset(a, b, sizeof a)
#define pb push_back
const int qq = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 10;
int num[qq];

int main() {
	int n, t;	scanf("%d%d", &n, &t);
	int ans = 0;
	for(int i = 1; i <= n; ++i) {
		scanf("%d", num + i);
	}
	for(int i = 1; i <= n; ++i) {
		t -= (86400 - num[i]);
		if(t <= 0) {
			ans = i;
			break;
		}
	}
	printf("%d\n", ans);
	return 0;
}

#include<bits/stdc++.h>

using namespace std;
#define LL long long
#define mst(a, b)	memset(a, b, sizeof a)
#define pb push_back
const int qq = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 10;

int main() {
	int n, k;	scanf("%d%d", &n, &k);
	int sum = 0;
	for(int x, i = 1; i <= n; ++i) {
		scanf("%d", &x);
		sum += x;
	}
	if(sum + n - 1 == k) {
		puts("YES");
	} else {
		puts("NO");
	}
	return 0;
}

可以改变两次要求可达的点对最大，分析可知我们把两个最大的圈连在一起即可达到最大

#include<bits/stdc++.h>

using namespace std;
#define LL long long
#define mst(a, b)	memset(a, b, sizeof a)
#define pb push_back
const int qq = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 10;
LL circle[qq];
bool vis[qq];
int a[qq];

int main() {
	int n;	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	int k = 0;
	for(int i = 1; i <= n; ++i) {
		if(!vis[i]) {
			int p = i;
			int cnt = 0;
			while(!vis[p]) {
				vis[p] = true;
				p = a[p];
				cnt++;
			}
			circle[k++] = cnt;
		}
	}
	sort(circle, circle + k);
	if(k == 1) {
		printf("%lld\n", circle[0] * circle[0]);
	} else {
		LL sum = 0;
		for(int i = 0; i < k - 2; ++i) {
			sum += circle[i] * circle[i];
		}
		sum += (circle[k - 1] + circle[k - 2]) * (circle[k - 1] + circle[k - 2]);
		printf("%lld\n", sum);
	}
	return 0;
}

#include<bits/stdc++.h>

using namespace std;
#define LL long long
#define mst(a, b)	memset(a, b, sizeof a)
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define pill pair<int, int>
const int qq = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 10;
priority_queue<LL, vector<LL>, greater<LL> > Q;

int main() {
	int n;	scanf("%d", &n);
	for(int x, i = 1; i <= n; ++i) {
		scanf("%d", &x);
		Q.push(x);
	}
	if(n == 1) {
		printf("0\n");
		return 0;
	}
	if(n % 2 == 0)	Q.push(0);
	LL ans = 0;
	while(Q.size() > 1) {
		LL tmp = Q.top();
		Q.pop();
		tmp += Q.top();
		Q.pop();
		tmp += Q.top();
		Q.pop();
		Q.push(tmp);
		ans += tmp;
	}
	printf("%lld\n", ans);
}