洛谷P2522 [HAOI2011]Problem b 题解

题目链接：P2522 [HAOI2011]Problem b

题意：对于给出的 \(n\) 个询问，每次求有多少个数对 \((x,y)\)，满足 \(a \le x \le b\)，\(c \le y \le d\)，且 \(\gcd(x,y) = k\)，\(\gcd(x,y)\) 函数为 \(x\) 和 \(y\) 的最大公约数。

一句话题意：

求 \[ \sum_{i=a}^{b}\sum_{j=c}^{d}[\gcd(i,j)=k] \] 根据容斥原理，不妨令 \[ F(n,m) = \sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)=k] \] 则答案为 \[ F(b,d)-F(b,c-1)-F(a-1,d)+F(a-1,c-1) \] 然后推推柿子 \[ \begin{aligned} &\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)=k] \\&=\sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}[\gcd(i,j)=1] \\&=\sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}\sum_{d\mid \gcd(i,j)}\mu(d) \end{aligned} \] 考虑变换求和顺序，得 \[ \sum_{d=1}^{\min\left(\left\lfloor\frac{n}{k}\right\rfloor,\left\lfloor\frac{m}{k}\right\rfloor\right)}\mu(d)\sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}[d\mid i]\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}[d\mid j] \] 根据 \(1\sim n\) 中 \(d\) 的倍数有 \(\left\lfloor\dfrac{n}{d}\right\rfloor\) 个，

以及 \(\left\lfloor\dfrac{n}{kd}\right\rfloor=\left\lfloor\frac{\left\lfloor\frac{n}{k}\right\rfloor}{d}\right\rfloor\)

可得 \[ \sum_{d=1}^{\min\left(\left\lfloor\frac{n}{k}\right\rfloor,\left\lfloor\frac{m}{k}\right\rfloor\right)}\mu(d)\left\lfloor\dfrac{n}{kd}\right\rfloor\left\lfloor\dfrac{m}{kd}\right\rfloor \] 然后数论分块就好了

时间复杂度 \(\mathcal{O}(N+Q\sqrt{n})\)

代码：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF 0x3f3f3f3f3f3f3f3f
namespace FastIO
{
    #define gc() readchar()
    #define pc(a) putchar(a)
    #define SIZ (int)(1e6+15)
    char buf1[SIZ],*p1,*p2;
    char readchar()
    {
        if(p1==p2)p1=buf1,p2=buf1+fread(buf1,1,SIZ,stdin);
        return p1==p2?EOF:*p1++;
    }
    template<typename T>void read(T &k)
    {
        char ch=gc();T x=0,f=1;
        while(!isdigit(ch)){if(ch=='-')f=-1;ch=gc();}
        while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        k=x*f;
    }
    template<typename T>void write(T k)
    {
        if(k<0){k=-k;pc('-');}
        static T stk[66];T top=0;
        do{stk[top++]=k%10,k/=10;}while(k);
        while(top){pc(stk[--top]+'0');}
    }
}using namespace FastIO;
#define N (int)(5e4+14)
int prime[N],pcnt,mu[N],sum[N];
bool ck[N];
void Mobius()
{
    mu[1]=1;
    for(int i=2; i<=N; i++)
    {
        if(!ck[i])
        {
            prime[++pcnt]=i;
            mu[i]=-1;
        }
        for(int j=1; j<=pcnt&&i*prime[j]<=N; j++)
        {
            int pos=i*prime[j];
            ck[pos]=1;
            if(i%prime[j])
            {
                mu[pos]=-mu[i];
            }else
            {
                mu[pos]=0;
                break;
            }
        }
    }
    for(int i=1; i<=N; i++)
        sum[i]+=sum[i-1]+mu[i];
}
int Q,a,b,c,d,k;
int solve(int n,int m)
{
    int res=0;
    n/=k;m/=k;
    for(int l=1,r; l<=min(n,m); l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        res+=(sum[r]-sum[l-1])*(n/l)*(m/l);
    }
    return res;
}
signed main()
{
    // ios::sync_with_stdio(0);
    // cin.tie(0);cout.tie(0);
    // freopen("check.in","r",stdin);
    // freopen("check.out","w",stdout);
    Mobius();read(Q);
    while(Q--)
    {
        read(a);read(b);read(c);read(d);read(k);
        write(solve(b,d)+solve(a-1,c-1)-solve(b,c-1)-solve(a-1,d));
        pc('\n');
    }
    return 0;
}