lightoj-1027Dangerous Maze（期望&概率DP）

You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.

If you choose the ith door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after xi minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.

Now you want to find the expected time to get out of the maze.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer n (1 ≤ n ≤ 100) denoting the number of doors. The next line contains nspace separated integers. If the ith integer (xi) is positive, you can assume that the ith door will take you out of maze after xi minutes. If it's negative, then the ith door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 10000.

Output

For each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result and q is the denominator of the result and they are relatively prime. See the samples for details.

Sample Input

3

1

1

2

-10 -3

3

3 -6 -9

Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

題意：

有n個門，每個門有權值，正數和負數，如果x[i] >= 0代表選擇i門後，x[i]分鐘後，帶你離開迷宮，如果x[i] < 0，代表-x[i]分鐘後，會回到起始位置，問你走出迷宮的期望步數

思路：

根據期望的定義（加權平均），我們可以首先列出一個步數期望方程，設走出迷宮的期望步數爲E，正數個數cnt1，負數個數cnt2,正數和sum1，負數和sum2

cnt1 + cnt2 = n

E = (sum1 / cnt1) * (cnt1 / n) + (sum2 / cnt2 + E) * (cnt2 / n)

E = (sum1/n) + (sum2 / n) + (cnt2 / n) * E

E =  ((sum1/n) + (sum2 / n) )  /  (cnt1 / n) 

∴E = (sum1 + sum2) / cnt1 

這個解方程其實沒必要寫的。。。 從而就可以得到E了，然後特判一下沒有正數就是沒有辦法出去的情況

int main()
{
    int t,cas = 1;cin >> t;
    while(t--){
        cin >> n;
        ll sum1 = 0,sum2 = 0;
        for(int i = 1;i <= n;i++)
            scanf("%lld",&a[i]);
        ll cnt1 = 0,cnt2 = 0;
        for(int i = 1;i <= n;i++){
            if(a[i] > 0) cnt1++,sum1 += a[i];
            else if(a[i] < 0) cnt2++,sum2 += a[i];
        }
        printf("Case %d: ",cas++);
        if(cnt1 == 0){
            cout << "inf" << endl;
        }else{
            sum2 = abs(sum2);
            ll fenmu = (sum1 + sum2);
            ll fenzi = cnt1;
            ll gcd = __gcd(fenmu,fenzi);
            printf("%lld/%lld\n",fenmu / gcd ,fenzi / gcd);
        }
    }
    return 0;
}