USACO 1.4.3 Arithmetic Progressions

/*
ID: tgh7281
LANG: C++
TASK: ariprog
*/
 
#include <iostream>
#include <fstream>
 
using namespace std;
bool status[250*250*2+1];
 
struct progression
{
    int a; //首项
    int b; //公差
}ariprog[10000];
 
bool check(int a,int b,int n);
int cmp(const void*a,const void*b);
 
int main()
{
    int i,j;
    ifstream fin("ariprog.in");
    ofstream fout("ariprog.out");
 
    memset(status,false,sizeof(status));
    int n,m;
    fin>>n>>m;
    for (i=0;i<=m;i++) //initial
    {
        for (j=i;j<=m;j++)
        {
            status[i*i+j*j]=true;
        }
    }
 
    int a,b,count=0,M=2*m*m;
 
    for (a=0;a<=M;a++) //枚举首项
    {
        if (!status[a]) continue;
        for (b=1;a+(n-1)*b<=M;b++) //枚举公差
        {
            if (check(a,b,n))
            {
                ariprog[count].a=a;
                ariprog[count].b=b;
                count++;
            }
        }
    }
 
    if (count)
    {
        qsort(ariprog,count,sizeof(ariprog[0]),cmp);
        for (i=0;i<count;i++) fout<<ariprog[i].a<<’ ‘<<ariprog[i].b<<endl;
    }
    else fout<<"NONE"<<endl;
 
    fin.close();
 
    fout.close();
 
    return 0;
}
 
bool check(int a,int b,int n)
{
    for (int i=1;i<n;i++)
        if (!status[a+i*b]) return false;
 
    return true;
}
 
int cmp(const void*a,const void*b)
{
    progression*c=(progression*)a;
    progression*d=(progression*)b;
 
    if (c->b==d->b) return c->a-d->a;
    return c->b-d->b;
}