Problem Description
Givenan integer N(0 ≤ N ≤ 10000), your task is to calculate N!
Input
OneN in one line, process to the end of file.
Output
Foreach N, output N! in one line.
Sample Input
1
2
3
Sample Output
1
2
6
Author
JGShining(极光炫影)
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高精度数,大数阶乘
类比十进制,模拟一个万进制的算法算法,参考:http://blog.csdn.net/lulipeng_cpp/article/details/7437641
用int 存数位,10000*100000<2^31,所以我每一位存了 100000,当然与 10000是类似的
*********/
Code:
#include<iostream> #include<stdio.h> #include<string.h> #include<cmath> using namespace std; /** 计算阶乘位数,顺便把POJ 1423 给A了 #define PI 3.141592653589793239 #define ee 2.7182818284590452354 int ans(int n){ return (int)((n*log10(n/ee)+log10(sqrt(2*n*PI))))+1; } */ int num[8000]; int main() { int n,i,j; while(cin>>n) { memset(num,0,sizeof(num)); num[0] = 1; for(i = 2;i<=n;i++) { for(j = 0;j<8000;j++) num[j]*=i; for(j = 0;j<8000;j++) { num[j+1] += num[j]/100000; num[j] %= 100000; } } int len = 8000; while(num[len] == 0) { len--; } cout<<num[len]; for(i = len-1;i>=0;i--) { printf("%.5d",num[i]); } printf("\n"); } return 0; }
时间: 2024-11-05 15:59:03