「Luogu3455」[POI2007]ZAP-Queries

「Luogu3455」[POI2007]ZAP-Queries

problem

Solution

题目要求

[sum_{x=1}^asum_{y=1}^b[gcd(a,b)=d] ]

设上式为(f(d))，构造(g(n)=sum_{n|d}f(d))，于是有

[g(n)=sum_{x=1}^asum_{y=1}^b[n|gcd(a,b)] ]

易得

[g(d)=lfloorfrac{a}{d} floorlfloorfrac{b}{d} floor ]

代入反演一下

[f(n)=sum_{n|d}mu(frac{d}{n})lfloorfrac{a}{d} floorlfloorfrac{b}{d} floor ]

令(t=frac{d}{n})，代入得

[f(n)=sum_{t=1}^frac{min(a,b)}{n}mu(t)lfloorfrac{a}{nt} floorlfloorfrac{b}{nt} floor ]

预处理(mu)的前缀和，整除分块即可

Code

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 50005
#define N 50000
using namespace std;
typedef long long ll;

template <typename T> void read(T &t)
{
    t=0;int f=0;char c=getchar();
    while(!isdigit(c)){f|=c=='-';c=getchar();}
    while(isdigit(c)){t=t*10+c-'0';c=getchar();}
    if(f)t=-t;
}

int n;
int pri[maxn],pcnt,nop[maxn],mu[maxn];
int a,b,k;

void GetPrime()
{
    mu[1]=1,nop[1]=1;
    for(register int i=2;i<=N;++i)
    {
        if(!nop[i])pri[++pcnt]=i,mu[i]=-1;
        for(register int j=1;j<=pcnt && i*pri[j]<=N;++j)
        {
            nop[i*pri[j]]=1;
            if(i%pri[j]==0)break;
            else mu[i*pri[j]]=-mu[i];
        }
    }
    for(register int i=1;i<=N;++i)mu[i]+=mu[i-1];
}

int Calc()
{
    int re=0,up=min(a,b)/k;
    for(register int l=1,r;l<=up;l=r+1)
    {
        r=min(a/(a/l),b/(b/l));
        re+=(mu[r]-mu[l-1])*(a/(l*k))*(b/(l*k));
    } 
    return re;
}

int main()
{
    GetPrime();
    read(n);
    while(n--)
    {
        read(a),read(b),read(k);
        printf("%d
",Calc());
    }
    return 0;
}