LightOJ 1027 A Dangerous Maze(期望)

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题意：n扇门，每扇所花时间为abs(x[i])，若x[i]为负则无法走出，为正则走出迷宫。求走出迷宫的期望时间。

思路：设cnt为x为正的门数。则一次走出迷宫的概率为 cnt / n，走出迷宫的期望次数为 n / cnt。走一次的平均时间为sum(abs[x[i]]) / n。则期望时间 ans = sum / n * n / cnt = sum / cnt。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 1e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//  freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    si(n);    int cnt = 0, ans = 0;    for(int i = 1; i <= n; i ++) {     si(m);    ans += abs(m);    if(m > 0) cnt ++;    }    printf("Case %d: ", ++ cas);    if(!cnt) { puts("inf"); continue; }    printf("%d/%d\n", ans / __gcd(ans, cnt), cnt / __gcd(ans, cnt));    }        return 0;}

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