bzoj 2820（Mobius）

传送门
一道没有真正意义上进行反演的“反演” 题。

#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int MAXN=1e7+2;
bool vis[MAXN];
int prime[MAXN/10],mu[MAXN],num=0,n,m;
ll f[MAXN];
inline int read() {
    int x=0;char c=getchar();
    while (c<'0'||c>'9') c=getchar();
    while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x;
}
inline void linear_shaker() {
    mu[1]=1,f[1]=0;
    memset(vis,false,sizeof(vis));
    for (register int i=2;iif (!vis[i]) prime[++num]=i,mu[i]=-1;
        for (int j=1;j<=num&&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<=num;++i)
        for (int j=1;j<=MAXN&&j*prime[i]for (register int i=2;i1];
}
inline ll cal(int n,int m) {
    ll ans=0;
    int t=min(n,m),last;
    for (int i=1;i<=t;i=last+1) {
        last=min(n/(n/i),m/(m/i));
        ans+=(f[last]-f[i-1])*(n/i)*(m/i);
    }
    return ans;
}
int main() {
//  freopen("bzoj 2820.in","r",stdin);
    int kase=read();
    linear_shaker();
    while (kase--) {
        n=read(),m=read();
        printf("%lld\n",cal(n,m));
    }
    return 0;
}