求n(10000以內)的階乘
阿新 • • 發佈:2019-02-03
#include<iostream>
#include<string>
#include<time.h>
#include<iomanip>
using namespace std;
#include<vector>
void cheng(vector<unsigned long long int> &result, int t)
{
int temp=0;
for (auto &x : result)
{
unsigned long long int q = (x*t + temp) /1000000000000000 ;
x = (x*t+temp) % 1000000000000000;
temp = q;
}
if (temp != 0) result.push_back(temp);
}
int main()
{
std::ios::sync_with_stdio(false);
// int a = clock();
vector<unsigned long long int> result;
result.push_back(1);
result.clear();
result.push_back(1);
int N;cin >> N;
for (int t = 2;t <= N;t++)
cheng(result, t);
cout << result.back();
for (auto it = result.rbegin() + 1;it != result.rend();it++)
cout << setw(15) << setfill('0') << *it;
cout << endl;
// int b = clock();
// cout << b - a << endl;
}