洛谷P3327 约数个数和

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
const int MAX_N = 50025;
int prime[MAX_N],pcnt,mu[MAX_N],summ[MAX_N],sd[MAX_N];
bool vis[MAX_N];
void sieve(int n)
{
    mu[1] = 1;
    for(int i = 2;i<n;++i)
    {
        if(!vis[i])
        {
            prime[pcnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0;j<pcnt&&prime[j]*i<n;j++)
        {
            int k = i *prime[j];
            vis[k] = true;
            if(i%prime[j]==0)
            {
                mu[k] = 0;
                break;
            }
            mu[k] = -mu[i];
        }
    }
    for(int i = 1;i<n;++i)
        summ[i] = summ[i-1] + mu[i];
    for(int i = 1;i<=n;++i)
    {
        ll res = 0;
        for(int l = 1,r;l<=i;l = r+1)
        {
            r = (i/(i/l));
            res+=1ll*(r-l+1)*1ll*(i/l);
        }
        sd[i] = res;
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    sieve(MAX_N);
    int n,m,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(n>m) swap(n,m);
        ll ans = 0;
        for(int l = 1,r;l<=n;l = r+1)
        {
            r = min(n/(n/l),m/(m/l));
            ans+=1ll*(summ[r]-summ[l-1])*1ll*sd[n/l]*1ll*sd[m/l];
        }
        printf("%lld\n",ans);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}
洛谷P3327 约数个数和

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