LightOJ 1027 A Dangerous Maze (数学期望)

题意：你面前有 n 个门，每次你可以选择任意一个进去，如果xi是正数，你将在xi后出去，如果xi是负数，那么xi后你将回来并且丢失所有记忆，问你出去的期望。

析：两种情况，第一种是直接出去，期望就是 1/n * xi

第二种是回来了，再出去 1/n*(-xi+E)，

然后就可以得到 E = sum / (n - cnt)。

sum是所有的数的绝对值的和，cnt是xi为负数的个数。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}


int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    int cnt = n;
    int sum = 0;
    for(int i = 0; i < n; ++i){
      scanf("%d", &m);
      sum += abs(m);
      if(m < 0)  --cnt;
    }
    printf("Case %d: ", kase);
    if(cnt == 0){
      printf("inf\n");
      continue;
    }
    int g = gcd(sum, cnt);
    printf("%d/%d\n", sum/g, cnt/g);
  }
  return 0;
}