HDU - 5072

HDU 5072

There are n people standing in a line. Each of them has a unique id number. Now the Ragnarok is coming. We should choose 3 people to defend the evil. As a group, the 3 people should be able to communicate. They are able to communicate if and only if their id numbers are pairwise coprime or pairwise not coprime. In other words, if their id numbers are a, b, c, then they can communicate if and only if (a,b)=(b,c)=(a,c)=1 or (a,b)≠1and(a,c)≠1and(b,c)≠1

, where (x, y) denotes the greatest common divisor of x and y. We want to know how many 3-people-groups can be chosen from the n people. Input

The first line contains an integer T (T ≤ 5), denoting the number of the test cases.

For each test case, the first line contains an integer n(3 ≤ n ≤ 10 ⁵), denoting the number of people. The next line contains n distinct integers a ₁, a ₂, . . . , a _n(1 ≤ a _i ≤ 10 ⁵) separated by a single space, where a _i stands for the id number of the i-th person.

Output For each test case, output the answer in a line.

这道题我一看以为是教我容斥三个数所以当场就懵逼了

原来这可以转换一下的

求两两不互质的三个数或者两两互质的三个数的总和等于所有的三元组个数减去其中两个数互质另两个不互质的三元组的个数。。。。。。

设一共有N个数，对每个数找出有a个数与它互质，那么两个数互质另两个不互质的三元组的个数为a*(N-a-1)/2，然后加加减减搞一搞。。。。。。

这里除以2是为了去重，设三元组a,b,c，a,b互质，b,c不互质，如果a,c不互质，那么a,b可互换，如果a,c互质，那么b,c可互换

哇，我好蠢啊

#include <iostream> #include <algorithm> #include <vector> #include <stack> #include <string.h> #include <cstdio> #include <map> #include <queue> #include <math.h> #include <cstring> #include <set> #define pb push_back #define fi first #define se second #define INF 0x3f3f3f3f using namespace std; typedef long long LL; const LL MOD=1e9+7; const int MAXN=100000+233; LL prime[MAXN],total; bool isprime[MAXN]; void make() { int m=MAXN-3; memset(isprime,true,sizeof(isprime)); isprime[0]=isprime[1]=false; total=0; for(int i=2;i<=m;i++) { if(isprime[i])prime[++total]=i; for(int j=1;j<=total&&prime[j]*i<=m;j++) { isprime[prime[j]*i]=false; if(i%prime[j]==0)break; } } } LL n,to; LL a[MAXN],fac[MAXN][33],b[MAXN]; LL z[MAXN]; bool vis[MAXN]; int main() { make(); int T; scanf("%d",&T); while(T--) { scanf("%lld",&n); memset(vis,false,sizeof(vis)); for(LL i=0;i<n;i++) { scanf("%lld",&a[i]); vis[a[i]]=true; } LL res=0; memset(b,0,sizeof(b)); memset(z,0,sizeof(z)); for(LL i=0;i<n;i++) { LL N=a[i]; for(int j=1;j<=total&&prime[j]*prime[j]<=N;j++) { LL p=prime[j]; if(N%p==0) { fac[i][b[i]++]=p; N/=p; } while(N%p==0)N/=p; } if(N>1) { fac[i][b[i]++]=N; } } for(LL i=1;i<=MAXN;i++) { for(LL j=i;j<=MAXN;j+=i) { if(vis[j]) { z[i]++; } } }//对于每个数，有多少个是他的倍数 to=(n)*(n-1)*(n-2)/6; for(int i=0;i<n;i++) { if(a[i]==1)continue; //printf("a[i]=%lld**********\n",a[i]); LL len=(1<<b[i]),s=0,r=0,f=1; for(int j=1;j<len;j++) { s=0;f=1; for(int k=0;k<b[i];k++) { if((1<<k)&j) { s++; f*=fac[i][k]; } } //printf("f=%d\n",f); if(s&1) { r+=z[f]-1; } else r-=z[f]-1; } res+=r*(n-r-1); } printf("%lld\n",to-res/2); } return 0; } 注：只是记个笔记，有错误的地方请大佬们指正