1 for(int i = 1; i <= m; ++ i) // 其中数组A存的即为大数相乘的结果
 2 {
 3     int b = 0; // 处理余数问题
 4     for(int j = 1; j <= len; ++ j)
 5     {
 6         int t = b + i*A[j];
 7         A[j] = t % 10;
 8         b = t / 10;
 9         if(b > 0 && j == len) len ++;
10     }
11 }

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <stack>
 7 #include <map>
 8 #include <queue>
 9 
10 using namespace std;
11 const int MAXN = 100005;
12 
13 int main()
14 {
15     int m, A[MAXN] = {1, 1}, len = 1;
16     scanf("%d", &m);
17 
18     for(int i = 1; i <= m; ++ i)
19     {
20         int b = 0;
21         for(int j = 1; j <= len; ++ j)
22         {
23             int t = i * A[j] + b;
24             A[j] = t % 10;
25             b = t / 10;
26             if (j == len && b > 0) ++len;
27         }
28     }
29 
30     for(int i = len; i >= 1; -- i)
31         printf("%d", A[i]);
32     printf("\n");
33     return 0;
34 }