Brave balloonists 求素因子的个数

最新推荐文章于 2020-11-02 22:25:29 发布

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最新推荐文章于 2020-11-02 22:25:29 发布

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原文链接：http://www.cnblogs.com/heqinghui/archive/2012/07/23/2605522.html

十个数学家乘坐热气球穿越赤道庆祝时，香槟瓶塞意外使氢气泄漏，气球开始下降。为延长他人生命，需决定一名牺牲者。通过每位数学家写下1到10000之间的整数并计算这些数乘积的所有正因数数量N的最后一位数来决定。

Description

Ten mathematicians are flying on a balloon over the Pacific ocean. When they are crossing the equator they decide to celebrate this event and open a bottle of champagne. Unfortunately, the cork makes a hole in the balloon. Hydrogen is leaking out and the balloon is descending now. Soon it will fall into the ocean and all the balloonists will be eaten by hungry sharks.
But not everything is lost yet. One of the balloonists can sacrifice himself jumping out, so that his friends would live a little longer. Only one problem still exists ¾ who is the one to get out. There is a fair way to solve this problem. First, each of them writes an integer a_i not less than 1 and not more than 10000. Then they calculate the magic number N that is the number of positive integer divisors of the product a₁*a₂*...*a₁₀. For example, the number of positive integer divisors of 6 is 4 (they are 1,2,3,6). The hero (a mathematician who will be thrown out) is determined according to the last digit of N. Your task is to find this digit.

Input

Input file contains ten numbers separated by a space.

Output

Output file should consist of a single digit from 0 to 9 - the last digit of N.

Sample Input

Sample Output

#include<stdio.h>
#include <iostream>
#include<cmath>
#include <algorithm>
using namespace std;
int save[10000];
int main()
{
	int a[10],i,ans,t,sum=1,k;
	for(k=0;k<10;k++)
	{
		scanf("%d",&a[k]);
		t=a[k];
		for(i=2;i<=t;i++)
		{
			while(t%i==0)
			{
				save[i]++;
				t=t/i;
			}
		}
	}
	for(i=0;i<=10000;i++)
		if(save[i]!=0)
		   sum*=save[i]+1;
	ans=sum%10;
	printf("%d\n",ans);
	return 0;
}

转载于:https://www.cnblogs.com/heqinghui/archive/2012/07/23/2605522.html