bzoj2820 YY的GCD

http://www.elijahqi.win/archives/1214
Description
神犇YY虐完数论后给傻×kAc出了一题给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对kAc这种
傻×必然不会了，于是向你来请教……多组输入
Input
第一行一个整数T 表述数据组数接下来T行，每行两个正整数，表示N, M
Output
T行，每行一个整数表示第i组数据的结果
Sample Input
2
10 10
100 100
Sample Output
30
2791
HINT

T = 10000

N, M <= 10000000

#include<cstdio>
#include<algorithm>
#define N 11000000
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (S==T){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while (ch<'0'||ch>'9') ch=gc();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc(); }
    return x;
}
int mu[N],T,not_prime[N],prime[1000000],g[N];
int main(){
    //freopen("bzoj2820.in","r",stdin);
    T=read();mu[1]=1;int top=0;
    for (int i=2;i<=N-1;++i){
        if (!not_prime[i]){
            mu[i]=-1;prime[++top]=i;
        }for (int j=1;prime[j]*i<=N-1;++j){
            not_prime[i*prime[j]]=true;
            if (i%prime[j]==0){
                mu[i*prime[j]]=0;break;
            }   mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=top;++i)
        for (int j=1;j*prime[i]<=N-1;j++) g[j*prime[i]]+=mu[j];
    for (int i=1;i<=N-1;++i) g[i]+=g[i-1];
    while (T--){
        int n=read(),m=read();long long ans=0;int tmp=min(n,m);int last=0;
        for (int i=1;i<=tmp;i=last+1){
            last=min(n/(n/i),m/(m/i));
            ans+=(long long)(n/i)*(m/i)*(g[last]-g[i-1]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}