排列组合Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5013 Accepted Submission(s): 2187
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 | We have carefully selected several similar problems for you: 1085 1171 1398 2082 2152 正数型母函数 代码实现: #include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; double a[15]; double b[15]; double num[15]; int n1[15],n2[15],v[15]; int main() { num[0]=num[1]=1; for(int i=1;i<=10;i++) num[i]=num[i-1]*i; int n,m; while(scanf("%d%d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) { scanf("%d",&n2[i]); n1[i]=0;v[i]=1; } memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); a[0]=1; for (int i=1;i<=n;i++) { memset(b,0,sizeof(b)); for (int j=n1[i];j<=n2[i]&&j*v[i]<=m;j++) for (int k=0;k<=m&&k+j*v[i]<=m;k++) b[k+j*v[i]]+=(a[k]/num[j]); memcpy(a,b,sizeof(b)); }
printf("%.0lf\n",a[m]*num[m]); } return 0; } |