# Southeastern European 2008 Sky Code 解题报告

http://www.cn210.com/onlinejudge/problemshow.php?pro_id=92

## Description

tancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to steal the spacecraft of Petru. There is only one problem - Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars from the Milky Way Galaxy. For breaking the system Stancu has to check each subset of four stars such that the only common divisor of their numbers is 1. Nasty, isn't it? Fortunately, Stancu has succeeded to limit the number of the interesting stars to N but, any way, the possible subsets of four stars can be too many. Help him to find their number and to decide if there is a chance to break the system.

## Input

For each test case on the first line the number N of interesting stars is given (1 ≤ N ≤ 10000). The second line of the test case contains the list of ID numbers of the interesting stars, separated by spaces. Each ID is a positive integer which is no greater than 10000. The input data terminate with the end of file.

## Output

For each test case the program should print one line with the number of subsets with the asked property.

## Sample Input

4
2 3 4 5
4
2 4 6 8
7
2 3 4 5 7 6 8


## Sample Output

1
0
34


res = C(4,s) - ΣC(4,k(n))(-1)^p(n)

#include <cstdio>
#include <iostream>
#include <cstring>
//#include <ctime>
using namespace std;

//Prime_Template_Var_Begin
const long MAXP = 10000;
long prime[MAXP] = {0},num_prime = 0;
int isNotPrime[MAXP] = {1, 1};
//Prime_Template_Var_End
int icCount = {0};
long long C4 = {0};
long countNum;
int main()
{
//freopen("d:\\b.in", "r", stdin);
//clock_t tim_rcd = clock();

long n, i, j, maxnum;
long count;
long long psn, res;
//Prime_Begin
for(i = 2 ; i <  MAXP ; i ++)
{
if(! isNotPrime[i])
prime[num_prime ++]=i;
for(j = 0 ; j < num_prime && i * prime[j] <  MAXP ; j ++)
{
isNotPrime[i * prime[j]] = 1;
if( !(i % prime[j]))
break;
}
}
//Prime_End
//Par_Count_Begin
for(i = 0; i < num_prime; i ++)
{
icCount[ prime[i] ] = 1;
for(j = 2; j < 10005 && j * prime[i] < 10005; j ++)
{
if(j % prime[i] && icCount[j])
icCount[j * prime[i]] = icCount[j] + 1;
}
}
for(j = 2; j < 10005; j ++)
if(icCount[j])
icCount[j] = (icCount[j] % 2)? 1: -1;
//Par_Count_End
//Count_C4_Begin
for(i = 4; i < 10005; i ++)
C4[i] = ((long long)((i) * (i - 1) / 2) * (long long)(i - 2) / 3) * (long long)(i - 3) / 4;
//Count_C4_End
while(scanf("%ld", &n) != EOF)
{
maxnum = 0;
memset(countNum, 0, sizeof(countNum));
for(i = 0; i < n; i ++)
{
scanf("%ld", &count);
maxnum = (maxnum > count)? maxnum: count;
countNum[count] ++;
for(j = 2; j * j <= count; j ++)
{
if(count % j == 0)
{
countNum[j] ++;
if(j * j != count)
countNum[count / j] ++;
}
}
}
if(n < 4)
{
printf("0\n");
continue;
}
psn = 0;
for(i = 2; i <= maxnum; i ++)
{
if(!icCount[i])
continue;
psn += icCount[i] * C4[countNum[i]];
}
res = C4[n];
printf("%lld\n", res - psn);
}

//cout<< (clock() - tim_rcd) * 1000 / CLK_TCK << endl;
return 0;
}