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：

2

10 10

100 100

输出样例#1：

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;

}