bzoj2880

打公式好麻烦　QAQ　　

为了节省时间去复习，原谅我引用一下别人的博客。。。http://blog.csdn.net/acdreamers/article/details/8542292

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#define re(i,l,r) for(int i=(l);i<=(r);i++)
#define rre(i,r,l) for(int i=(r);i>=(l);i--)
using namespace std;
typedef long long LL;
int prime[10000010],prime_tot,mu[10000010],g[10000010],sum[10000010];
bool bo[10000010];
LL solve(int n,int m)
{
    LL ret=0;
    if(n>m)swap(n,m);
    int i=1,xia;
    while(i<=n)
    {
        xia=min(n/(n/i),m/(m/i));
        ret+=1LL*(n/i)*(m/i)*(sum[xia]-sum[i-1]);
        i=xia+1;
    }
    return ret;
}
int main()
{
    mu[1]=1;
    re(i,2,10000000)
    {
        if(!bo[i])prime[++prime_tot]=i,mu[i]=-1,g[i]=1;
        for(int j=1;j<=prime_tot&&i*prime[j]<=10000000;j++)
        {
            bo[i*prime[j]]=1;
            if(i%prime[j]==0){mu[i*prime[j]]=0,g[i*prime[j]]=mu[i];break;}
            else mu[i*prime[j]]=-mu[i],g[i*prime[j]]=mu[i]-g[i];
        }
    }
    re(i,1,10000000)sum[i]=sum[i-1]+g[i];
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;scanf("%d%d",&n,&m);
        printf("%lld
",solve(n,m));
    }
    return 0;
}