lightoj 1027 期望公式

假设期望为E 如果走死路门就相当于徒劳走回来又是一个循环
所以E = (a1+a2+a3+….am+mE)/n+(b1+b2+b3+b4)/n
(n-m)E = sum(all value);
ai为死路门的时间 bj为每个出口出去的时间

#include<cstdio>#include<cstring>#include<algorithm>#include<cstring>#include<iostream>using namespace std;int gcd(int a,int b){    while(b)    {        int t = a%b;        a = b;        b = t;    }    return a;}int main(){    int t,i1 = 1;    scanf("%d",&t);    while(t--)    {        int n,cnt = 0,ans = 0;        scanf("%d",&n);        while(n--)        {            int a;            scanf("%d",&a);            if(a>0)cnt++;            ans+=abs(a);        }        printf("Case %d: ",i1);        i1++;        if(!cnt){printf("inf\n");continue;}        int pre = gcd(ans,cnt);        printf("%d/%d\n",ans/pre,cnt/pre);    }    return 0;}