题解 | N的阶乘

#include <iostream>
using namespace std;

void bigNumMul(int n){
	int i,j,k,rear=1,m,count=0;
	long long result[512]={0},temp,d;
	k=n;
	while (k != 0){
		k=k/10;
		count++;
	}
	d=0;
	result[1]=1;
	m=(n*count)/8 + 1;
	for(i=2;i<=n;i++){
		for(j=1;j<=m;j++){
			temp = result[j]*i + d;
			result[j] = temp % 100000000;
			d = temp / 100000000;  // 不能用result[j] 已经改变值 
		}
	} 
	for(i=m;i>=1;i--){
		if(result[i] != 0){
			rear = i;
			break;
		}
	}
	cout<<result[rear];   // 8位以下，不用补零。只用输出数组第一个数据
	for(i=rear-1;i>=1;i--){
		if(result[i]>9999999){
			cout<<result[i];
			continue;
		}
		if(result[i]>999999){
			cout<<"0"<<result[i];
			continue;
		}
		if(result[i]>99999){
			cout<<"00"<<result[i];
			continue;
		}
		if(result[i]>9999){
			cout<<"000"<<result[i];
			continue;
		}
		if(result[i]>999){
			cout<<"0000"<<result[i];
			continue;
		}
		if(result[i]>99){
			cout<<"00000"<<result[i];
			continue;
		}
		if(result[i]>9){
			cout<<"000000"<<result[i];
			continue;
		}
		cout<<"0000000"<<result[i];
	}
}


int main(){
	int n;
	while(cin>>n){
		bigNumMul(n);
        cout<<endl;
	}
	return 0;
}