排列组合[HDU1521]

排列组合

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1736    Accepted Submission(s): 726

 

Problem Description
有n种物品，并且知道每种物品的数量。要求从中选出m件物品的排列数。例如有两种物品A,B，并且数量都是1，从中选2件物品，则排列有"AB","BA"两种。

 

 

 

Input
每组输入数据有两行，第一行是二个数n,m(1<=m,n<=10)，表示物品数，第二行有n个数，分别表示这n件物品的数量。
 

 

Output
对应每组数据输出排列数。(任何运算不会超出2^31的范围)
 

 

Sample Input
2 2
1 1
 

 

Sample Output
2
 

 

Author
xhd
 

 

Recommend
xhd

 

A typical application of exponential generating function.The i-th good's generating function is (1+x/1!+x^2/2!+......+x^C[i]/C[i]!).In the product of the n functions,the coefficient of the item whose index of x is n multiplies n! is the finally answer.

 

#include<stdio.h>
#include<string.h>
double p[15],frac[15],tmp[15];
int c[15];
int n,m;
int main()
{
    int n,m,i,j,k;
    frac[0]=1;
    for (i=1;i<12;i++) frac[i]=frac[i-1]*i;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (i=1;i<=n;i++) scanf("%d",&c[i]);
        memset(p,0,sizeof(p));
        for (i=0;i<=c[1];i++) p[i]=1.0/frac[i];
        for (i=2;i<=n;i++)
        {
            memset(tmp,0,sizeof(tmp));
            for (j=0;j<=m;j++)
                for (k=0;k<=c[i];k++)
                    tmp[j+k]+=p[j]*(1.0/frac[k]);
            for (j=0;j<=m;j++) p[j]=tmp[j];
        }
        printf("%.0lf
",p[m]*frac[m]);
    }
    return 0;
}