N!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 68606 Accepted Submission(s): 19629
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
题意：还需要解释么。。。1W以内的阶乘运算。
题解：一开始想着是不是打个表，后面MLE了几次，哭死，最后也懒得本地打表，索性把打表撤了一次一次算，耗时有点长，不过还是A了。
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>using namespace std;
struct BigInt
{const static int mod = 10000;const static int DLEN = 4;int a[10000],len;BigInt(){memset(a,0,sizeof(a));len = 1;}BigInt(int v){memset(a,0,sizeof(a));len = 0;do{a[len++] = v%mod;v /= mod;}while(v);}BigInt(const char s[]){memset(a,0,sizeof(a));int L = strlen(s);len = L/DLEN;if(L%DLEN)len++;int index = 0;for(int i = L-1; i >= 0; i -= DLEN){int t = 0;int k = i - DLEN + 1;if(k < 0)k = 0;for(int j = k; j <= i; j++)t = t*10 + s[j] - '0';a[index++] = t;}}BigInt operator +(const BigInt &b)const{BigInt res;res.len = max(len,b.len);for(int i = 0; i <= res.len; i++)res.a[i] = 0;for(int i = 0; i < res.len; i++){res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);res.a[i+1] += res.a[i]/mod;res.a[i] %= mod;}if(res.a[res.len] > 0)res.len++;return res;}BigInt operator *(const BigInt &b)const{BigInt res;for(int i = 0; i < len; i++){int up = 0;for(int j = 0; j < b.len; j++){int temp = a[i]*b.a[j] + res.a[i+j] + up;res.a[i+j] = temp%mod;up = temp/mod;}if(up != 0)res.a[i + b.len] = up;}res.len = len + b.len;while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;return res;}void output(){printf("%d",a[len-1]);for(int i = len-2; i >=0 ; i--)printf("%04d",a[i]);printf("\n");}
};
BigInt a,temp;
int main(int argc, const char * argv[])
{a=BigInt(1);int n;while(cin >> n){a=BigInt(1);for(int i=2;i<=n;i++){temp=BigInt(i);a=a*temp;}a.output();}return 0;
}