hdu5970

真是怀疑当初合肥赛区怎么考这么差的……

首先根据辗转相除法可知f(i,j)=f(i+j*k,j)

于是我们可以先于处理出f(i,j) (j<=666,i<=j)，当确定i,j时c也确定

（x=gcd(i,j)）可见，当确定了i,j,k后，后面p的求和就是一个等差数列的求和

c是logm级的所以总复杂度就是O(T*m^2*logm)

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
int f[670][670],c[670][670],n,m,p;
int get(int x,int y,int &f,int &c)
{
    int t; c=0;
    while (y)
    {
        c++;
        t=x%y; x=y; y=t;
    }
    f=c*x*x;
}

int main()
{
    int cas;
    scanf("%d",&cas);
    for (int j=1; j<=666; j++)
        for (int i=1; i<=j; i++)
            get(i,j,f[i][j],c[i][j]);
    while (cas--)
    {
        scanf("%d%d%d",&n,&m,&p);
        ll ans=0;
        for (int j=1; j<=m; j++)
            for (int i=1; i<=j&&i<=n; i++)
                for (int k=0; k<c[i][j]; k++)
                {
                    if (i+k*j>n) break;
                    ll b=(i+j*k)*j/f[i][j];
                    ll d=c[i][j]*j*j/f[i][j];
                    ll t=(n-(i+j*k))/(c[i][j]*j)+1;
                    ans=(ans+b*t%p+(t-1)*t/2%p*d%p)%p;
                }
        printf("%lld
",ans);
    }
}