[HDOJ1261]字串数

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1261

解题思路很好想，设一共有n=n1+n2+...nk个字符，分别出现n1,n2,...nk次，则组合数有n!/(n1!n2!...nk!)种

难点在大数运算上

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <cmath>
 7 #include <queue>
 8 #include <map>
 9 #include <set>
10 #include <stack>
11 #include <list>
12 #include <vector>
13
14 using namespace std;
15
16 int ans[500];
17 int has[27];
18 int n, s;
19
20 void division(int n, int m) {
21     int c = 0;
22     for(int i = 0; i < 500; i++) {
23         c = ans[i] * n + c;
24         ans[i] = c % 10;
25         c /= 10;
26     }
27     c = 0;
28     for(int i = 499; i >= 0; i--) {
29         c = ans[i] + c * 10;
30         ans[i] = c / m;
31         c %= m;
32     }
33 }
34
35 int main() {
36     // freopen("in", "r", stdin);
37     while(~scanf("%d", &n) && n) {
38         s = 0;
39         memset(ans, 0, sizeof(ans));
40         ans[0] = 1;
41         for(int i = 0; i < n; i++) {
42             scanf("%d", &has[i]);
43             for(int j = 1; j <= has[i]; j++) {
44                 division(s+j, j);
45             }
46             s += has[i];
47         }
48         int i;
49         for(i = 499; i >= 0; i--) {
50             if(ans[i]) {
51                 break;
52             }
53         }
54         for(int j = i; j >= 0; j--) {
55             printf("%d", ans[j]);
56         }
57         printf("\n");
58     }
59 }