N!

Problem Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

 

 

Input

One N in one line, process to the end of file.

 

 

Output

For each N, output N! in one line.

 

 

Sample Input

1
2
3

 

 

Sample Output

1
2
6

 

 

Author

JGShining（极光炫影）

分析： 要注意时间限制，特别当位数很大时，要进行的乘法操作也会很大！解决方法是每三位进行一次乘法操作！

参考答案：

 #include <iostream> #include <cmath> #include <string> using namespace std; void init(char v[]) { memset(v, 0, sizeof(v)); v[0] = '1'; } void multi(char v[], int mul) { int val, wadd, j; wadd = 0; for (j = 0; j < strlen(v); ++j) { if (v[j]=='0'&&wadd == 0) continue; val = (v[j]-'0')*mul+wadd; wadd = val/10; v[j] = val%10+'0'; } while (wadd != 0) { val = wadd%10; wadd = wadd/10; v[j++] = val+'0'; } v[j++] = '/0'; } void multiQurick(char v[], int mul) { int val, wadd = 0, j; int l = strlen(v), sub; for (j = 0; j <= l-1; ++j) { sub = l-1-j; if (sub == 0) { val = (v[j]-'0'); val = val*mul+ wadd; v[j] = val%10+'0'; wadd = val/10; } else if (sub == 1) { val = (v[j]-'0')+ (v[j+1]-'0')*10; val = val*mul+ wadd; v[j] = val%10+'0'; val = val/10; v[++j] = val%10+'0'; wadd = val/10; } else { val = (v[j]-'0')+ (v[j+1]-'0')*10 +(v[j+2]-'0')*100; val = val*mul+ wadd; v[j] = val%10+'0'; val = val/10; v[++j] = val%10+'0'; val = val/10; v[++j] = val%10+'0'; wadd = val/10; } } while (wadd != 0) { val = wadd%10; wadd = wadd/10; v[j++] = val+'0'; } v[j++] = '/0'; } void revert(char v[]) { int l = strlen(v), i; char tmp; for (i = 0; i < l/2; ++i) { tmp = v[i]; v[i] = v[l-i-1]; v[l-i-1] = tmp; } } int main() { int n, i; char res[100000]; while(cin>>n) { init(res); for (i = 1; i <= n; ++i) { multiQurick(res, i); } revert(res); cout<<res<<endl; } return 0; }