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  • hdu 5663 Hillan and the girl 莫比乌斯反演

    Hillan and the girl

    Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)


    Problem Description
    “WTF! While everyone has his girl(gay) friend, I only have my keyboard!” Tired of watching others' affair, Hillan burst into scream, which made him decide not to hold it back.
    “All right, I am giving you a question. If you answer correctly, I will be your girl friend.” After listening to Hillan, Girl replied, “What is the value of ni=1mj=1f(i,j), where f(i,j)=0 if gcd(i,j) is a square number and f(i,j)=1 if gcd(i,j) is not a square number(gcd(i,j) means the greatest common divisor of x and y)?”
    But Hillan didn't have enough Intelligence Quotient to give the right answer. So he turn to you for help.
     
    Input
    The first line contains an integer T(1T10,000)——The number of the test cases.
    For each test case, the only line contains two integers n,m(1n,m10,000,000) with a white space separated.
     
    Output
    For each test case, the only line contains a integer that is the answer.
     
    Sample Input
    2 1 2333333 10 10
     
    Sample Output
    0 33
    Hint
    In the first test case, obviously $fleft(i,j ight)$ always equals to 0, because $i$ always equals to 1 and $gcdleft(i,j ight)$ is always a square number(always equals to 1).
     
    Source
                      min(n,m) min(n/k,m/k)
    思路:首先推到∑     ∑ mu(d)  * [n/k/d] * [m/k/d];  k为完全平方数;
          k=1   d=1
       令T=k*d;
       可得:
          min(n,m)                    
          ∑   [n/T] * [m/T]  ∑   mu(T/k)  ;
          T          k|T
          令gg数组等于 ∑   mu(T/k)  相当于原来的mu函数;
                 k|T
          和原来一样分块即可;
    #include<bits/stdc++.h>
    using namespace std;
    #define ll __int64
    #define esp 0.00000000001
    #define pi 4*atan(1)
    const int N=1e7+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
    int mu[N], p[N], np[N], cnt, sum[N];
    ll gg[N];
    void init() {
        mu[1]=1;
        for(int i=2; i<N; ++i) {
            if(!np[i]) p[++cnt]=i, mu[i]=-1;
            for(int j=1; j<=cnt && i*p[j]<N; ++j) {
                int t=i*p[j];
                np[t]=1;
                if(i%p[j]==0) { mu[t]=0; break; }
                mu[t]=-mu[i];
            }
        }
        for(int i=1;i*i<N;i++)
        {
            for(int t=i*i;t<N;t+=(i*i))
            sum[t]+=mu[t/i/i];
        }
        for(int i=1;i<N;i++)
        gg[i]=gg[i-1]+sum[i];
    
    }
    ll getans(ll b,ll d)
    {
        if(b>d)swap(b,d);
        ll ans=0;
        for(ll L=1,R=0;L<=b;L=R+1)
        {
            R=min(b/(b/L),d/(d/L));
            ans+=(b/L)*(d/L)*(gg[R]-gg[L-1]);
        }
        return ans;
    }
    int main()
    {
        int T;
        init();
        scanf("%d",&T);
        while(T--)
        {
            ll b,d;
            scanf("%I64d%I64d",&b,&d);
            printf("%I64d
    ",(b*d)-getans(b,d));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5790129.html
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