NYoj 416 数论的一个小知识

话说这道题是昨天月赛的题目，是一个简单的数论题。可悲的是，当时竟然不知道这个结论，比赛过后，才知道这道题有多么的水。昨天的比赛，那就是一个悲剧。剩下一个半小时左右，连一次也没提交。。。。。。。悲催。。。。

题目：

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 ai 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.

输入 The first line of input contains a number c(0<c<=1000) giving the number of cases that follow. each case contains ten numbers separated by a space.
输出 Output each case of a single digit from 0 to 9 - the last digit of N 样例输入

1
1
2
6
1
3
1
1
1
1
1

样例输出

9

ac代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <string.h>
using namespace std;
int aa[1500];
int kk;
bool isprime(int n){
for(int j=2;j<=sqrt(n);++j){
if(n%j==0)
return false;
}
return true;
}
void prime(){
kk=0;
for(int i=2;i<=10000;++i){

if(isprime(i)){
aa[kk++]=i;
}

}
//cout<<kk<<"  "<<aa[kk-1]<<endl;
}
int main(){
prime();
int count;
scanf("%d",&count);
while(count--){
int n=10,x;
int sum[10000];
memset(sum,0,sizeof(sum));
while(n--){
scanf("%d",&x);
for(int j=0;j<kk;++j){
if(x%aa[j]==0){
while(x%aa[j]==0){
x/=aa[j];
sum[j]++;
}
}
}

}
int num=1;
for(int i=0;i<10000;++i){
if(sum[i]){
//printf("%d\n",sum[i]);
num=(num%10)*((sum[i]+1)%10)%10;
}
}
printf("%d\n",num);
}
return 0;
}