hdu 1261 字串数 排列组合

字串数

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3505    Accepted Submission(s): 855

Problem Description

一个A和两个B一共可以组成三种字符串:"ABB","BAB","BBA".
给定若干字母和它们相应的个数,计算一共可以组成多少个不同的字符串.

 

Input

每组测试数据分两行,第一行为n(1<=n<=26),表示不同字母的个数,第二行为n个数A1,A2,...,An(1<=Ai<=12),表示每种字母的个数.测试数据以n=0为结束.

 

Output

对于每一组测试数据,输出一个m,表示一共有多少种字符串.

 

Sample Input

21 232 2 20

 

Sample Output

390

 可以轻易推出公式 ：（n1+n2+n3+...nn)!/(n1!+n2!+...+nn!);

代码：

#include <stdio.h>#include <string.h>#define SIZE 30typedef long long ll ;int d[SIZE] ;int ans[1000] , f[15];void multiply(int c){ans[0] = ans[1] = 1 ;for(int i = 2 ; i <= c ; ++i){int r = 0 ;for(int j = 1 ; j <= ans[0] ; ++j){ans[j] *= i ; ans[j] += r ;r = ans[j]/10 ; ans[j] %= 10  ; }if(r != 0){while(r){ans[ans[0]+1] += r%10 ;ans[0] = ans[0]+1 ;r /= 10 ;}}}}void divide(int n){for(int i = 0 ; i < n ; ++i){if(d[i] == 1)continue ;ll r = 0 ;for(int j = ans[0] ; j > 0 ; --j){r = r*10 + ans[j] ;ans[j] = (int)(r/f[d[i]]) ;r %= f[d[i]] ;}int j = ans[0] ;while(!ans[j--]) ;ans[0] = j+1 ;}}int main(){int n ;f[0] = f[1] = 1 ;for(int i = 2 ; i < 15 ; ++i)f[i] = f[i-1]*i ;while(scanf("%d",&n) && n){int c = 0; memset(ans,0,sizeof(ans)) ;for(int i = 0 ; i < n ; ++i){scanf("%d",&d[i]) ;c += d[i] ;}multiply(c) ; divide(n) ;for(int i = ans[0] ; i > 0 ; --i)printf("%d",ans[i]) ;puts("") ;}return 0 ;}/*2612 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12答案： 43040483600756137098280735739634397715612462348749865063078010246084385786829462878395555377796951146973463603476517994060086036378690699890922398850719933177533763394328048206442072300423203375709346966736476550911025693768450467040156990933760328149015001020695298485434314813181979093662554695180376557841893905005438946222775672233517571336196144202713162201212368185472256427516516825563136000000000000000000000000*/

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