ACM: 概率DP 动态规划题 poj 1322

Chocolate

Description

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

"Green, orange, brown, red...", colorful sugar-coated shell maybeis the most attractive feature of ACM chocolate. How many colorshave you ever seen? Nowadays, it's said that the ACM chooses from apalette of twenty-four colors to paint their delicious candybits.

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

Now, here comes the problem, if there are C colors of ACMchocolates in the package (colors are distributed evenly), after Nchocolates are taken from the package, what's the probability thatthere is exactly M chocolates on the table? Would you please writea program to figure it out?

Input

The input file for this problemcontains 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 singlezero.

Output

The output should be one realnumber per line, shows the probability for each case, round tothree decimal places.

Sample Input

5 100 20

Sample Output

0.625 

题意: 一共有c种巧克力, 每种的个数都是一样多并且是足够多, 现在从包里面拿出n次巧克力一次一个,

     当桌面上有2个相同的时候就吃掉, 现在问你在桌面上出现m个巧克力的概率.

解题思路:

    1. 状态dp[i][j]: 前i次操作时, 桌面上出现j个巧克力的概率. (i+j)为奇数是dp[i][j]=0;

    2. 问题：状态方程:dp[i][j] = dp[i-1][j-1]*(c-j+1.0)/c + dp[i-1][j+1]*（j+1.0)/c;     (原来错误的方程：dp[i][j] = dp[i-1][j-1]*(c-j)/c + dp[i-1][j+1]*j/c;)

      谢谢指出此错误的博客朋友.

                p = (c-j+1.0)/c: 当前抽取出来的概率不相同.

                q = (j+1.0)/c: 下一个抽取出来的概率相同.

           因为这样的结果都是桌面上剩下: j-1个巧克力.

        3. 郁闷的问题: 从网上别人的思考发现, 当抽取次数越来越大的时候, 概率已经是趋于平衡.

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 105

int c, n, m;
double dp[MAX*10][MAX];

int main()
{
// freopen("input.txt","r",stdin);
 while(scanf("%d",&c) != EOF)
 {
  if(c == 0) break;
  scanf("%d %d",&n,&m);
  if(m > c || m > n || (m+n)%2 != 0)
  {
   printf("0.000\n");
   continue;
  }

  memset(dp,0,sizeof(dp));
  if(n > 1000) n = 1000+n%2;
  dp[0][0] = 1;
  for(int i = 1; i <= n; ++i)
  {
   for(int j = 0; j <= c; ++j)
   {
    if( (i+j)%2 != 0 ) continue;
    dp[i][j] = dp[i-1][j-1]*(c-j+1.0)/c + dp[i-1][j+1]*(j+1.0)/c;
   }
  }
  printf("%.3lf\n",dp[n][m]);
 }
 return 0;
}