【概率DP】 LightOJ 1027 A Dangerous Maze

题意：N个门，接下有N个数

正数X能花费X时间逃出迷宫，负数Y表示-Y时间重新选，选每个门的概率一样

求花费的时间的期望

ANS= P1*X+(1-P2)*(Y+ ANS)；

X表示正数平均值

Y表示abs(负数)平均值

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <deque>
#include <map>
#define cler(arr, val) memset(arr, val, sizeof(arr))
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long LL;
const int MAXN = 302020;
const int MAXM = 1602020;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const double eps = 1e-4;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int gcd(int x,int y)
{
if(y==0) return x;
else return gcd(y,x%y);
}
int main()
{
int cas=1,t,n,a[200];
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
#endif
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int sum1,sum2,x1,x2;
sum1=sum2=x1=x2=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]>0) sum1+=a[i],x1++;
else sum2-=a[i],x2++;
}
int ans=sum1+sum2;
int g=gcd(ans,x1);
if(sum1==0)
printf("Case %d: inf\n",cas++);
else printf("Case %d: %d/%d\n",cas++,ans/g,x1/g);
}
return 0;
}