POJ2603

                                                                                                                                   Brave balloonists

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4721 Accepted: 2013

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 exists¾ 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

1261311111

Sample Output

9思想：本题输入十个数，求出这十个数相乘最后所得数的因子个数，再对因子个数取个位，p1p2…pr为素数，      因子个数为(1+a1)*(1+a2)*(1+a3)*...*(1+an)对于每个ai，有ai+1种取法。解题代码：#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;#define N 10005bool ispri[N];int pri[N];int get_pri[N];void getpri(){   memset(ispri, true, sizeof(ispri));   ispri[0] = ispri[1] = 0;   int t = 0;   for(int i = 2 ; i <= N ; i ++)   {       if(ispri[i])       {           pri[++t] = i;           for(int j = i * i; j <= N ; j += i)                ispri[j] = false;       }   }   pri[0] = 0;}void solve(int n){    for(int i = 1; pri[i] * pri[i] < n ; i ++)    {        if(n % pri[i] == 0)        {            while(n % pri[i] == 0)            {                get_pri[pri[i]]++;                n /= pri[i];            }        }    }    if(n > 1)        get_pri[n]++;}int main(){//freopen("in.txt","r",stdin);    getpri();    int a;    for(int i = 0; i < 10; i ++)    {        scanf("%d",&a);        solve(a);    }    int ans = 1;    for(int i = 1 ; i < N ; i ++)    {        if(get_pri[i])            ans = ans * (get_pri[i] + 1) % 10;    }    printf("%d\n",ans);    }