Problem Description
1 5 1 3 9 10 2
4
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e5 + 5;
int mob[MAX], p[MAX];
int a[MAX], num[MAX], has[MAX], cp[MAX];
bool prime[MAX];
int ma, n;
int Mobius()
{
int pnum = 0;
memset(prime, true, sizeof(prime));
mob[1] = 1;
for(int i = 2; i < MAX; i++)
{
if(prime[i])
{
p[pnum ++] = i;
mob[i] = -1;
}
for(int j = 0; j < pnum && i * p[j] < MAX; j++)
{
prime[i * p[j]] = false;
if(i % p[j] == 0)
{
mob[i * p[j]] = 0;
break;
}
mob[i * p[j]] = -mob[i];
}
}
}
ll cal()
{
ll all = (ll) n * (n - 1) * (n - 2) / 6;
memset(cp, 0, sizeof(cp));
memset(num, 0, sizeof(num));
for(int i = 1; i <= ma; i++)
{
for(int j = i; j <= ma; j += i)
num[i] += has[j];
for(int j = i; j <= ma; j += i)
cp[j] += mob[i] * num[i];
}
ll no = 0;
for(int i = 0; i < n; i++)
if(a[i] != 1)
no += (ll) cp[a[i]] * (n - cp[a[i]] - 1);
return all - no / 2;
}
int main()
{
Mobius();
int T;
scanf("%d", &T);
while(T--)
{
memset(has, 0, sizeof(has));
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%d", &a[i]);
has[a[i]] ++;
ma = max(a[i], ma);
}
printf("%I64d\n", cal());
}
}版权声明:本文为博主原创文章,未经博主允许不得转载。
HDU 5072 Coprime (莫比乌斯反演+容斥+同色三角形)
原文地址:http://blog.csdn.net/tc_to_top/article/details/47784863