spoj GCDMAT

题目大意：给定 $a,b,c,d$,求 $\sum_{i=a}^b\sum_{j=c}^dgcd(i,j)$

$$\begin{align} \sum_{i=1}^n\sum_{j=1}^mgcd(i,j) & = \sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}[gcd(i,j) == 1] \\ & = \sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}\sum_{g|i,j}\mu(g) \\ & = \sum_{d=1}^nd\sum_{g=1}^{\lfloor\frac nd\rfloor}\mu(g)\lfloor\frac n{gd}\rfloor \lfloor\frac m{gd}\rfloor\\ \end{align}$$

然后就随便搞了嘛QAQ
大水怪，嘿嘿嘿

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<ctime>
#include<cmath>
#include<cstring>
#include<string>
#define LL long long
#define N 50002
#define M 1000000007
using namespace std;

int tot,T,n,m,a,b,c,d;
int miu[N],prime[N],v[N];

void get_prime()
{
    miu[1]=1;
    for (int i=2;i<N;i++)
    {
        if (!v[i]) miu[prime[tot++]=i]=-1;
        for (int j=0;j<tot&&i*prime[j]<N;j++)
        {
            v[i*prime[j]]=true;
            if (i%prime[j]==0) break;
            miu[i*prime[j]]=-miu[i];
        }
    }
    for (int i=2;i<N;i++) (miu[i]+=miu[i-1])%=M;
}

LL sum(int n,int m)
{
    LL ret=0;
    for (int i=1,j;i<=n;i=j+1)
    {
        j=min(m/(m/i),n/(n/i));
        (ret+=(LL)(miu[j]-miu[i-1])*(n/i)*(m/i))%=M;
    }
    return ret;
}

LL cal(int n,int m)
{
    if (!n||!m) return 0;
    LL ret=0;
    if (n>m) swap(n,m);
    for (int i=1,j;i<=n;i=j+1)
    {
        j=min(m/(m/i),n/(n/i));
        (ret+=(LL)(j-i+1)*(i+j)/2%M*sum(n/i,m/i))%=M;
    }
    return ret;
}

int main()
{
    scanf("%d",&T);
    scanf("%d%d",&n,&m);
    get_prime();
    while (T--)
    {
        scanf("%d%d%d%d",&a,&b,&c,&d);
        cout<<(cal(c,d)+cal(a-1,b-1)-cal(c,b-1)-cal(a-1,d)+M+M)%M<<endl;
    }

    return 0;
}