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

123

 

 

Sample Output

126

 

 

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;}