Codeforces Round #428 (Div. 2)

A

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;

int main(){
	int n, k;	scanf("%d%d", &n, &k);
	int cnt = 0;
	for(int i = 0; i < n; ++i) {
		int x;	scanf("%d", &x);
		cnt += x;
		if(cnt >= 8)	k -= 8, cnt -= 8;
		else	k -= cnt, cnt = 0;
		if(k <= 0) {
			printf("%d\n", i + 1);
			return 0;
		}
	}
	puts("-1");
	return 0;
}

B

题意：给出n行，每行有8个座位， {1, 2} {3 4} {4, 5} {5, 6} {7, 8} 一行中这些位置算相邻，给出k个不同部队的士兵，要求不同部队的士兵部能坐相邻的位置，问能否达到这个目的

昨晚- - 我也被叉掉了、

就很尴尬、

题解的思路很清晰、比我的要清晰多了QAQ

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;
int has[5], cnt[5];

int main(){
	int n, k;	scanf("%d%d", &n, &k);
	has[2] = n * 2, has[4] = n;
	for(int i = 0; i < k; ++i) {
		int x;	scanf("%d", &x);
		while(x >= 3) {
			if(has[4] > 0) {
				has[4]--, x -= 4;
			} else if(has[2] > 0) {
				has[2]--, x -= 2;
			} else {
				return puts("NO"), 0;
			}
		}
		if(x > 0)	cnt[x]++;
	}
	while(cnt[2] > 0) {
		if(has[2] > 0) {
			has[2]--, cnt[2]--;
		} else if(has[4] > 0) {
			cnt[2]--, has[4]--, has[1]++;
		} else {
			cnt[2]--, cnt[1] += 2;
		}
	}
	if(cnt[1] > has[1] + has[2] + has[4] * 2) {
		puts("NO");
	} else {
		puts("YES");
	}
	return 0;
}

C

题意：有n个节点n - 1条边，一头马在结点1，这头马有个习惯，只会走之前没有走过的结点，问最后深度的期望值

思路：只要注意到马在的当前节点有多少个子结点，它前往任何一个子结点的概率就是到达当前结点的概率 * (1 / 子结点数量)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>

using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define mst(a, b)	memset(a, b, sizeof a)
#define REP(i, x, n)	for(int i = x; i <= n; ++i)
const int MOD = 1e9 + 7;
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;
int dep[qq], num[qq];
vector<int> G[qq];
double ans = 0;
void Dfs(int u, int fa, int dis, double babl) {
	dep[u] = dis;
	int len = G[u].size();
	for(int i = 0; i < len; ++i) {
		int v = G[u][i];
		if(v == fa)	continue;
		if(u != 1)	Dfs(v, u, dis + 1, babl * ((1.0) / (len - 1)));
		else	Dfs(v, u, dis + 1, babl * ((1.0) / (len)));
		num[u] += num[v];
	}
	num[u] = num[u] + 1;
	if(num[u] == 1) {
		ans += dep[u] * 1.0 * babl;
	}
}

int main(){
	int n;	scanf("%d", &n);
	for(int i = 1; i < n; ++i) {
		int a, b;	scanf("%d%d", &a, &b);
		G[a].pb(b), G[b].pb(a);
	}
	Dfs(1, -1, 0, 1);
	printf("%.12lf\n", ans);
	return 0;
}