POJ 2603

Brave balloonists

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4934		Accepted: 2074

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

1
2
6
1
3
1
1
1
1
1

Sample Output

9

Source

Ural State University collegiate programming contest 2000

题意：

求十个数的乘积的因子个数；

解题思路：把十个数的结果进行因式分解得：s=a1^p1*a2^p2*...

则在0到p1内取任意个a1,0到p2内取任意个a2,所以最后的约数个数为：（p1+1）*(p2+1)*...%10

 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#define maxn  10007

using namespace std;
int num[11];
int ans[maxn];
bool prime[maxn];
void slove(int value)
{
        for(int i=2;i*i<=value;i++)
        {
          while(value%i==0){
            ans[i]++;
            value=value/i;
          }
        }
        if(value!=1) ans[value]+=1;
}
int main()
{
   // freopen("in.txt","r",stdin);
    memset(ans,0,sizeof(ans));
    for(int i=0;i<10;i++) scanf("%d",&num[i]);
    for(int i=0;i<10;i++) slove(num[i]);
    long long res=1;
    for(int i=2;i<maxn;i++)
        res*=(ans[i]+1),res=res%10;
        res=res%10;
        printf("%lld\n",res);
        return 0;
}