POI2007_zap 莫比乌斯反演

题意：http://hzwer.com/4205.html

 

同hdu1695

 

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
#define LL long long
#define MMX 50010
int mu[MMX],msum[MMX];
LL n;
bool check[MMX];
int prime[MMX];

void Moblus()                //莫比乌斯反演，mu[i]为莫比乌斯函数
{
    memset(check,false,sizeof(check));
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i <= MMX; i++)
    {
        if( !check[i] )
        {
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < tot; j++)
        {
            if(i * prime[j] > MMX) break;
            check[i * prime[j]] = true;
            if( i % prime[j] == 0)
            {
                mu[i * prime[j]] = 0;
                break;
            }
            else
            {
                mu[i * prime[j]] = -mu[i];
            }
        }
    }
    msum[1]=mu[1];                //求和，求G的时候用
    for (int i=2;i<=MMX;i++)
        msum[i]=msum[i-1]+mu[i];
}

LL G(int n,int m)           //G(x,y)表示有多少组x<=n，y<=m，且x,y互质  （(x,y)和(y,x)算两组）
{
    LL ans = 0;
    if(n > m)   swap(n,m);
    for(int i = 1, la = 0; i <= n; i = la+1)
    {
        la = min(n/(n/i),m/(m/i));
        ans += (LL)(msum[la] - msum[i-1])*(n/i)*(m/i);      //事先预处理：msum[n]=SUM(mu[1..n])
    }
    return ans;
}

int main()
{
    int T;
    cin>>T;
    Moblus();
    for (int zy=1; zy<=T; zy++)
    {
        int a,b,c,d,k;
        //cin>>b>>d>>k;
        scanf("%d%d%d\n",&b,&d,&k);
        if (b>d) swap(b,d);     //assume b<d
        b=b/k;
        d=d/k;

        LL ans1 = G(b,d);

        printf("%lld\n",ans1);
    }
}

 

1020

root

1003

Accepted

564K

3819MS

G++

1.62K

2014-11-16 18:08:01