P2257 YY的GCD

题面

https://www.luogu.org/problem/P2257

题解

#include<cstdio>
#include<iostream>
#define ri register int
#define N 10000050
#define mxN 10000000
#define LL long long

using namespace std;
int T,n,m;
int miu[N],g[N],p[N],pfk[N],prime[N/10];
LL sumg[N];

int main(){
  scanf("%d",&T);
  pfk[1]=1;miu[1]=1;
  ri cnt=0;

  for (ri i=2;i<=mxN;i++) {
    if (!p[i]) {
      prime[++cnt]=i;
      pfk[i]=i;
      miu[i]=-1;
    }
    for (ri j=1;prime[j]*i<=mxN && j<=cnt;j++) {
      ri x=prime[j]*i;
      p[x]=prime[j];
      pfk[x]=prime[j];
      if (i%prime[j]==0) {
        pfk[x]*=pfk[i];
        break;
      }
    }
  }
  for (ri i=2;i<=mxN;i++) if (p[i]) miu[i]=miu[pfk[i]]*miu[i/pfk[i]];

  for (ri j=1;j<=cnt;j++)
    for (ri p=prime[j],i=1;i*p<=mxN;i++) g[i*p]+=miu[i];
  for (ri i=1;i<=mxN;i++) sumg[i]=sumg[i-1]+g[i];

  while (T--) {
    scanf("%d %d",&n,&m);
    int p=1,q=1;    
    long long sum=0;
    ri i;
    for(int l=1,r;l<=min(n,m);l=r+1)
        {
            r=min(n/(n/l),m/(m/l));
            sum+=1ll*(n/l)*(m/l)*(sumg[r]-sumg[l-1]);
        }
    //for (ri i=1;i<=n && i<=m;i++) sum+=g[i]*(n/i)*(m/i);
    printf("%lld
",sum);
  }
  return 0;
}