LightOJ 1027 - A Dangerous Maze （期望）

题意:

迷宫n<=100扇门,每个门上面有个数字,要么是正数,要么是负数,选择门的概率都相同且随机
正数表示x秒后出去了,负数表示x秒后原地不变,求出去的期望

分析:

尝试写dp,failed!  贴个概率分析吧
设能走出去的门的个数为cnt，总的个数为n，那么一次走出去的概率为cnt/n，走一次的用的平均时间是sum(t)/n，期望的次数为n/cnt。时间期望是：(sum(t)/n)∗(n/cnt)=sum(t)/cnt

代码:

////  Created by TaoSama on 2015-11-02//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;struct Frac {    int p, q;    Frac(int a = 0, int b = 1): p(a), q(b) {        int gcd = __gcd(p, q);        p /= gcd; q /= gcd;    }    Frac operator* (const Frac& f) const {        return Frac(p * f.p, q * f.q);    }    Frac operator+ (const Frac& f) const {        return Frac(p * f.q + f.p * q, q * f.q);    }    bool operator< (const Frac& f) const {        return p * f.q < f.p * q;    }    void print() {printf("%d/%d\n", p, q);}};int n, x;int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        int sum = 0, cnt = 0;        for(int i = 1; i <= n; ++i) {            int x; scanf("%d", &x);            if(x > 0) ++cnt, sum += x;            else sum += -x;        }        if(!cnt) printf("Case %d: inf\n", ++kase);        else {            Frac ans = Frac(sum, cnt);            printf("Case %d: %d/%d\n", ++kase, ans.p, ans.q);        }    }    return 0;}