zoukankan      html  css  js  c++  java
  • LightOj-1027 A Dangerous Maze(期望)

    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

    题解:给你n扇门,每扇门都有一个数,k代表从这扇门走k分钟出去,-k代表从这扇门走-k分钟回到原点,让你求出去的期望;

    设期望为d 则有(以样例3位例子):d=(3+6+d+9+d)/3;

    参考代码为:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int T,n,a[110],sum,cnt;
     4 
     5 int gcd(int a,int b){ return b==0? a:gcd(b,a%b); }
     6 
     7 int main()
     8 {
     9     scanf("%d",&T);
    10     for(int k=1;k<=T;k++)
    11     {
    12         sum=0;cnt=0;
    13         scanf("%d",&n);
    14         for(int i=1;i<=n;i++)
    15         {
    16             scanf("%d",a+i),sum+=abs(a[i]);
    17             if(a[i]<0) cnt++;
    18         }
    19         if(cnt==n) printf("Case %d: inf
    ",k);
    20         else
    21         {
    22             cnt=n-cnt;
    23             if(cnt>sum) swap(cnt,sum);
    24             int num=gcd(sum,cnt);
    25             //int num= __gcd(sum,cnt);
    26             printf("Case %d: %d/%d
    ",k,sum/num,cnt/num);
    27         }
    28     }
    29     
    30     return 0;
    31 }
    View Code

       

  • 相关阅读:
    在线整数序列百科全书
    非常完整的线性DP及记忆化搜索讲义
    洛谷P2858 奶牛零食 题解 区间DP入门题
    HDU3394 Railway 题解(边双连通分量)
    POJ1144 Network 题解 点双连通分量(求割点数量)
    LibreOJ6279. 数列分块入门 3 题解
    LibreOJ 6278. 数列分块入门 2 题解
    LibreOJ 6277. 数列分块入门 1 题解
    洛谷P1020 导弹拦截 题解 LIS扩展题 Dilworth定理
    CF1272E. Nearest Opposite Parity 题解 广度优先搜索
  • 原文地址:https://www.cnblogs.com/csushl/p/9483730.html
Copyright © 2011-2022 走看看