ZOJ1363 Chocolate

Chocolate

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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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

2100年，ACM牌巧克力将风靡全球。

题意

“绿的，橘红的，棕色的，红的…”，彩色的糖衣可能是ACM巧克力最吸引人的地方。你一共见过多少种颜色？现在，据说ACM公司从一个24种颜色的调色板中选择颜色来装饰他们的美味巧克力。

有一天，Sandy用一大包有五种颜色的巧克力玩了一个游戏。每次他从包里拿出一粒巧克力并把它放在桌上。如果有桌上有两粒相同颜色的巧克力，他就把他们吃掉。他惊奇的发现大多数时候桌上都有2到3粒巧克力。

如果ACM巧克力有C(1≤C≤100)种颜色，在拿出了N(1≤N≤1000000)粒巧克力之后在桌上恰有M(1≤M≤1000000)粒的概率是多少？

分析

如果N不是那么大的话，我们是很容易用动态规划来解决此题，状态转移方程就是

P_i+1,k=P_i,k-1*(C-k-1)/C+P_i,k+1*(k+1)/C

其中P_i,k表示拿出了i粒巧克力后桌上剩余M粒的概率（当然还要考虑一些边界情况）。但是现在N可以达到1000000，如果直接动态规划肯定是要超时的。

这个题的标准算法是使用生成函数。也就是说把“桌上有m块巧克力”转化成“有m种巧克力取了奇数块，其余的都取偶数块的取法”。所以就可以列出生成函数是，所以总的取法数就是xⁿ的系数乘以n!和C(c,m)，而概率就是总的取法数除以cⁿ，然后通过进一步的代数分析来化简解决。这种方法当然是很优秀的，复杂度是O(c²)。但是由于这道题的精度要求很低，迭代的方法也是可以达到目的的，而且复杂度也接近O(c²)。

这道题里不会出现极大或极小的概率，一般来说这种情况下的收敛是比较快的。我们可以不断的计算P的值，当它的变化不足以影响结果时就停止计算。当然这道题里的收敛是分奇偶的（显然桌上剩余的巧克力数和拿出的巧克力数是同奇偶的），所以不能比较P_i和P_i-1，而要比较P_i和P_i-2，只要看到P_i和P_i-2差距小于一个定值（比如1e-5），而且i和N同奇偶，就可以停止动态规划，因为此时继续规划下去所产生的不同已经不可能影响到要输出的结果。经过实验发现，最大的数据也只在几百次迭代中就稳定了，这样就将效率大大提高，满足了题目的要求。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=1010;
double f[N][N];int n,m,c;
int main(){
    while(~scanf("%d",&c)&&c){
        scanf("%d%d",&n,&m);
        if(!n&&!m){puts("1.000");continue;}
        if(m>n||m>c||(n+m)&1){puts("0.000");continue;}
        if(n>1000) n=1000+(n&1);
        memset(f,0,sizeof f);f[1][1]=1;
        for(int i=2;i<=n;i++){
            for(int j=0;j<=min(i,c);j++){
                if(i+j&1){
                    f[i][j]=0;continue;
                } 
                if(j>0) f[i][j]+=f[i-1][j-1]*(1-(double)(j-1)/(double)c);
                if(j<i) f[i][j]+=f[i-1][j+1]*(double)(j+1)/(double)c;
            }
        }
        printf("%.3lf
",f[n][m]);
    }
    return 0;
}