YY的GCD 数学

题目描述

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

输入输出格式

输入格式：

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

输出格式：

T行，每行一个整数表示第i组数据的结果

输入输出样例

输入样例#1：
复制

2
10 10
100 100

输出样例#1： 复制

30
2791

说明

T = 10000

N, M <= 10000000

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
    int x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

bool vis[10000002];
int mu[10000002];
ll sum[10000002];
int p[10000002];
int f[10000001];
int tot;
void init() {
    mu[1] = 1;
    for (int i = 2; i <= 10000000; i++) {
        if (!vis[i]) {
            p[++tot] = i; mu[i] = -1;
        }
        for (int j = 1; j <= tot; j++) {
            if (i*p[j] > 10000000)break;
            vis[i*p[j]] = 1;
            if (i%p[j] == 0) {
                mu[i*p[j]] = 0; break;
            }
            else {
                mu[i*p[j]] = -mu[i];
            }
        }
    }
    for (int i = 1; i <= tot; i++) {
        for (int j = 1; j*p[i] <= 10000000; j++) {
            f[j*p[i]] += mu[j];
        }
    }
    for (int i = 1; i <= 10000000; i++)
        sum[i] = sum[i - 1] + 1ll * f[i];
}

int main()
{
    //	ios::sync_with_stdio(0);
    init();
    int T = rd();
    while (T--) {
        int N = rd(), M = rd();
        ll ans = 0;
        for (int l = 1, r; l <= min(N, M); l = r + 1) {
            r = min(N / (N / l), M / (M / l));
            ans += 1ll * (sum[r] - sum[l - 1])*(N / l)*(M / l);
        }
        printf("%lld\n", ans * 1ll);
    }
    return 0;
}