hdu 5072 Coprime (容斥)

Posted 2020-12-01 uid001

大意: 给定长$n$的序列$a$, 元素互不相同, 求有多少个三元组$(x,y,z)$满足两两互质或两两不互质.

 

考虑计算不合法情况. 若互质连一条白边, 不互质连一条黑边, 那么一个不合法的三元环一定有两个角是异色的, 合法三元环三个角都是同色的, 所以只要数出异色角即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘
‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



const int N = 1e5+10;
int n,a[N],s[N],vis[N];
int p[100],cnt;
ll ret;

void dfs(int d, int num, int z) {
	if (d>cnt) ret+=z*s[num];
	else dfs(d+1,num,z),dfs(d+1,num*p[d],-z);
}

ll solve(int x) {
	cnt = 0;
	for (int i=2; i*i<=x; ++i) if (x%i==0) {
		p[++cnt] = i;
		while (x%i==0) x/=i;
	}
	if (x>1) p[++cnt] = x;
	ret = 0, dfs(1,1,1);
	return ret;
}

void work() {
	scanf("%d",&n);
	REP(i,1,n) scanf("%d",a+i),vis[a[i]]=1;
	REP(i,1,N-1) {
		s[i] = 0;
		for (int j=i; j<N; j+=i) s[i]+=vis[j];
	}
	ll ans = 0;
	REP(i,1,n) { 
		if (a[i]!=1) {
			ll A = solve(a[i]), B = n-A-1;
			ans += A*B;
		}
		vis[a[i]] = 0;
	}
	printf("%lld
",(ll)n*(n-1)*(n-2)/6-ans/2);
}

int main() {
	int t;
	scanf("%d",&t);
	while (t--) work();
}