【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: 2820
    Language: C++
    Result: Accepted
    Time:4164 ms
    Memory: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;
}