bzoj 2820

subsubsection{例题3}
href{http://www.lydsy.com/JudgeOnline/problem.php?id=2820}{BZOJ 2820 YY的GCD}
题目大意：求有多少数对((x,y))满足$xin left[ 1,n ight] ,yin left[ 1,m ight] (满足)(x,y)(为质数 做法： par 首先这个题目和上一个题目不一样的地方是他需要一个特殊的转化 ext{令} k=&min(n,m);\ ans=&sum_{p}^ksum_{i=1}^nsum_{j=1}^mleft[ (i,j)=p ight] \ =&sum_{p}^ksum_{d=1}^k mu(d)lfloor frac{n}{pd} floor lfloor frac{m}{pd} floor \ ext{令}T=&pd\ ans=&sum_{T=1}^{k}lfloor frac{n}{T} floor lfloor frac{m}{T} floor sum_{p|T}^{k}mu(frac{T}{p})\ ext{令}F(k)=&sum_{p|T}^kmu(frac{T}{p})\ ext{则}ans=&sum_{T=1}^kF(k)lfloor frac{n}{T} floor lfloor frac{m}{T} floor \ egin{align} ext{令} k=&min(n,m);\ ans=&sum_{p}^ksum_{i=1}^nsum_{j=1}^mleft[ (i,j)=p ight] \ =&sum_{p}^ksum_{d=1}^k mu(d)lfloor frac{n}{pd} floor lfloor frac{m}{pd} floor \ ext{令}T=&pd\ ans=&sum_{T=1}^{k}lfloor frac{n}{T} floor lfloor frac{m}{T} floor sum_{p|T}^{k}mu(frac{T}{p})\ ext{令}F(k)=&sum_{p|T}^kmu(frac{T}{p})\ ext{则}ans=&sum_{T=1}^kF(k)lfloor frac{n}{T} floor lfloor frac{m}{T} floor \ end{align} 线性筛素数的时候对)F(k)$前缀和处理
然后就转变为和例二 ef{2}一样的做法，枚举除法的取值了
egin{lstlisting}[language={[ANSI]C}]

#include<iostream>
#include<cstdio>
#include<cmath>
#define N 10000000
#define ll long long 
using namespace std;
 
bool not_prime[N];
ll prime[N];
ll sum[N];
ll mu[N];
ll tot;
 
void Mu(int n){
    mu[1]=1;
    for(int i=2;i<=n;i++){
        if(!not_prime[i]){
            prime[++tot]=i;
            mu[i]=-1;
        }
        for(int j=1;prime[j]*i<=n;j++){
            not_prime[prime[j]*i]=1;
            if(i%prime[j]==0){
                mu[prime[j]*i]=0;
                break;
            }
            mu[prime[j]*i]=-mu[i];
        }
    }
    for(int i=1;i<=tot;++i)
        for(int j=1;j*prime[i]<=n;++j)
            sum[j*prime[i]]+=(ll)mu[j];
    for(int i=1;i<=n;++i)
        sum[i]+=(ll)sum[i-1];
}
 
ll ans(int n,int m){
    if(n>m)swap(n,m);
    int last,i;ll re=0;
    for(i=1;i<=n;i=last+1){
        last=min(n/(n/i),m/(m/i));
        re+=(ll)(n/i)*(m/i)*(sum[last]-sum[i-1]);
    }   
    return re;
}
 
int main(){
    Mu(N);
    int T;
    int a,b;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&a,&b);
        ll Ans=ans(a,b);
        printf("%lld
",Ans);
    }
    return 0;
}

end{lstlisting}
这已经是第n次被long long卡一个小时以上了