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$x\in \left[ 1,n\right] ,y\in \left[ 1,m \right] \(满足\)(x,y)\(为质数 做法： \par 首先这个题目和上一个题目不一样的地方是他需要一个特殊的转化 \text{令} k=&min(n,m);\\ ans=&\sum_{p}^k\sum_{i=1}^n\sum_{j=1}^m\left[ (i,j)=p\right] \\ =&\sum_{p}^k\sum_{d=1}^k \mu(d)\lfloor \frac{n}{pd}\rfloor \lfloor \frac{m}{pd}\rfloor \\ \text{令}T=&pd\\ ans=&\sum_{T=1}^{k}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \sum_{p|T}^{k}\mu(\frac{T}{p})\\ \text{令}F(k)=&\sum_{p|T}^k\mu(\frac{T}{p})\\ \text{则}ans=&\sum_{T=1}^kF(k)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \\ \begin{align} \text{令} k=&min(n,m);\\ ans=&\sum_{p}^k\sum_{i=1}^n\sum_{j=1}^m\left[ (i,j)=p\right] \\ =&\sum_{p}^k\sum_{d=1}^k \mu(d)\lfloor \frac{n}{pd}\rfloor \lfloor \frac{m}{pd}\rfloor \\ \text{令}T=&pd\\ ans=&\sum_{T=1}^{k}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \sum_{p|T}^{k}\mu(\frac{T}{p})\\ \text{令}F(k)=&\sum_{p|T}^k\mu(\frac{T}{p})\\ \text{则}ans=&\sum_{T=1}^kF(k)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \\ \end{align} 线性筛素数的时候对\)F(k)$前缀和处理\
然后就转变为和例二\ref{2}一样的做法，枚举除法的取值了\
\begin{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\n",Ans);
    }
    return 0;
}

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