【莫比乌斯反演】BZOJ2920-YY的GCD

【题目大意】

给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对。

【思路】

太神了这道题……蒟蒻只能放放题解：戳，明早再过来看看还会不会推导过程……

实用的结论：

 

嗯……

 

/**************************************************************Problem: 2820Language: C++Result: AcceptedTime:4164 msMemory:196600 kb
****************************************************************/#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x7fffffff;
const int MAXN=10000000+50;
typedef long long ll;
int miu[MAXN],g[MAXN],prime[MAXN],pnum=0;
ll sum[MAXN];
int N,M;void get_miu(int maxn)
{miu[1]=1;g[1]=0;sum[1]=sum[0]=0;for (int i=2;i<maxn;i++) miu[i]=-INF;for (int i=2;i<maxn;i++){if (miu[i]==-INF){miu[i]=-1;prime[++pnum]=i;g[i]=1;}for (int j=1;j<=pnum;j++){if (i*prime[j]>=maxn) break;if (i%prime[j]==0) {miu[i*prime[j]]=0;g[i*prime[j]]=miu[i];}else{miu[i*prime[j]]=-miu[i];g[i*prime[j]]=miu[i]-g[i];} }sum[i]=sum[i-1]+g[i];}
} void get_ans()
{ll ans=0; scanf("%d%d",&N,&M);if (N>M) swap(N,M);int pos;for (int t=1;t<=N;t=pos+1){pos=min(N/(N/t),M/(M/t));ans+=(ll)(sum[pos]-sum[t-1])*(N/t)*(M/t);}printf("%lld\n",ans);
}int main()
{get_miu(MAXN);int T;scanf("%d",&T);while (T--) get_ans();return 0;
}