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;elsek -= 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)));elseDfs(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;}