[BZOJ 2820]YY的GCD

[BZOJ 2820]YY的GCD

题目

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

OUTPUT

30

2791

解题报告

反演首题（说好NOIP不考反演的QAQ，最近考了这么多QAQ）

设$nleqslant m$

$$sum_{isprime(p)} sum_{a=1}^{n} sum_{b=1}^{m}gcd(a,b)==p$$

$$sum_{isprime(p)} sum_{a=1}^{left lfloor frac{n}{p} ight floor} sum_{b=1}^{left lfloor frac{m}{p} ight floor}gcd(a,b)==1$$

$$sum_{isprime(p)}  sum_{a=1}^{left lfloor frac{n}{p} ight floor} sum_{b=1}^{left lfloor frac{m}{p} ight floor} sum_{dmid gcd(a,b)} mu (d)$$

$$sum_{isprime(p)}  sum_{a=1}^{left lfloor frac{n}{p} ight floor} sum_{b=1}^{left lfloor frac{m}{p} ight floor} sum_{dmid awedge dmid b} mu (d)$$

$$sum_{isprime(p)}sum_{d=1}^{left lfloor frac{n}{p} ight floor}mu (d)left lfloor frac{n}{pd} ight floor left lfloor frac{m}{pd} ight floor$$

设$k=pd$

$$sum_{k=1}^{n}sum_{isprime(p)wedge pmid k}mu (frac{k}{p})left lfloor frac{n}{k} ight floor left lfloor frac{m}{k} ight floor$$

$$sum_{k=1}^{n}F(k)left lfloor frac{n}{k} ight floor left lfloor frac{m}{k} ight floor$$

woc，我tm写了点啥

然后就线性筛一下，可以GG

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
inline int read(){
    int sum(0);
    char ch(getchar());
    for(;ch<'0'||ch>'9';ch=getchar());
    for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar());
    return sum;
}
typedef long long L;
int t,n,m;
bool isnotprime[10000005];
int prime[10000005],mu[10000005],num;
L f[10000005];
inline void play_table(){
    mu[1]=1;
    for(int i=2;i<=10000000;++i){
        if(!isnotprime[i]){
            prime[++num]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=num&&i*prime[j]<=10000000;++j){
            isnotprime[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }
            else
                mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=num;++i){
        int p(prime[i]);
        for(int j=1;j*p<=10000000;++j)
            f[j*p]+=mu[j];
    }
    for(int i=1;i<=10000000;++i)
        f[i]+=f[i-1];
}
inline int gg(){
    freopen("YYnoGCD.in","r",stdin);
    freopen("YYnoGCD.out","w",stdout);
    play_table();
    t=read();
    while(t--){
        L ans(0);
        n=read(),m=read();
        if(n>m)
            swap(n,m);
        for(int i=1,j;i<=n;i=j+1){
            j=min(n/(n/i),m/(m/i));
            ans+=(f[j]-f[i-1])*(n/i)*(m/i);
        }
        printf("%lld
",ans);
    }
    return 0;
}
int K(gg());
int main(){;}