POJ 1322 Chocolate

Chocolate

Time Limit: 2000MS Memory Limit: 65536KTotal 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 20

Sample Output

0.625 

Source

Beijing 2002

 

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，颜色相同的吃掉，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即可，因为很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 1010using 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\n");            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\n",dp[n][m]);    }    return 0;}