zoukankan      html  css  js  c++  java
  • POJ 1322 Chocolate

    Chocolate
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 8245   Accepted: 2186   Special Judge

    Description

    In 2100, ACM chocolate will be one of the favorite foods in the world. 

    "Green, orange, brown, red...", colorful sugar-coated shell maybe is the most attractive feature of ACM chocolate. How many colors have you ever seen? Nowadays, it's said that the ACM chooses from a palette of twenty-four colors to paint their delicious candy bits. 

    One day, Sandy played a game on a big package of ACM chocolates which contains five colors (green, orange, brown, red and yellow). Each time he took one chocolate from the package and placed it on the table. If there were two chocolates of the same color on the table, he ate both of them. He found a quite interesting thing that in most of the time there were always 2 or 3 chocolates on the table. 

    Now, here comes the problem, if there are C colors of ACM chocolates in the package (colors are distributed evenly), after N chocolates are taken from the package, what's the probability that there is exactly M chocolates on the table?

    Would you please write a program to figure it out? 

    Input

    The input file for this problem contains several test cases, one per line. 

    For each case, there are three non-negative integers: C (C <= 100), N and M (N, M <= 1000000). 

    The input is terminated by a line containing a single zero. 

    Output

    The output should be one real number per line, shows the probability for each case, round to three decimal places.

    Sample Input

    5 100 2
    
    0
    

    Sample Output

    0.625 
    

    Source

    题意:C种颜色的巧克力在桶中,从里面依次拿出n个巧克力,颜色同样的吃掉。求最后剩下m个巧克力的概率

    当n>1000 时候,考虑奇偶性取1000或1001就可以,由于非常大的时候概率会趋于稳定,至于奇数时取1001 偶数

    时取1000有些不解

    #include <iostream>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <cstdlib>
    #include <cstdio>
    #define N 1010
    using namespace std;
    double dp[N][110];
    int main()
    {
        int c,n,m;
        while(scanf("%d",&c)!=EOF)
        {
            if(c==0)
            {
                break;
            }
            scanf("%d %d",&n,&m);
            if(m>c||m>n||(n-m)%2)
            {
                printf("0.000
    ");
                continue;
            }
            if(n>1000)
            {
                n = 1000+n%2;
            }
            memset(dp,0,sizeof(dp));
            dp[0][0] = 1;
            dp[1][1] = 1;
            for(int i=1;i<=n;i++)
            {
                for(int j=0;j<=i&&j<=c;j++)
                {
                    if(j-1>=0)
                    {
                        dp[i][j] = dp[i-1][j-1]*(double)(c-j+1)/(double)c;
                    }
                    dp[i][j] += dp[i-1][j+1]*(double)(j+1)/(double)c;
                }
            }
            printf("%.3lf
    ",dp[n][m]);
        }
        return 0;
    }
    
  • 相关阅读:
    System.nanoTime()的使用
    只为高效、流畅开发 —— 码云企业版 3.0 倾情上线
    不自律的人,下场都很惨
    刘德:小米已投89家生态链企业 有品要做百亿电商平台(本质上是是利用了小米的大火炉的余热,但也有反向的正面作用)
    英雄无敌手游(战争纪元云中城,还可以骑龙,绝美)
    openFrameworks 是一个旨在助力你进行开创性工作的开源 C++ 工具箱(是很多其它类库的组合)
    Core开发-MVC 使用dotnet 命令创建Controller和View
    Ant Table组件
    web性能优化
    scss + react + webpack + es6
  • 原文地址:https://www.cnblogs.com/yangykaifa/p/7063397.html
Copyright © 2011-2022 走看看