usaco Arithmetic Progressions

预处理全部q2+p2有20000多个。

枚举任意两个的差，并以这个差扩展找到连续的有几个

为了避免重复浪费时间，只需判断无法向左扩展的一对数，具体见代码。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in1.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

const int M = 255;

int Set[M*M*2];
int bisquares[M*M];
int n,m;

struct result
{
    int a,b;
    result(int a=0,int b=0):a(a),b(b){}
    bool operator < (const result &ant) const
    {
        if(b!=ant.b)return b<ant.b;
        else return a<ant.a;
    }
}res[11000];
int init()
{
    memset(Set,0,sizeof(Set));

    for(int i=0;i<=m;i++)
    {
        for(int j=0;j<=m;j++)
            Set[i*i+j*j]=1;
    }
    int num=0;
    for(int i=0;i<=m*m*2;i++)
    {
        if(Set[i])
            bisquares[num++]=i;
    }
    return num;
}

int check(int a,int b)
{
    while(a+b<=m*m*2&&Set[a+b])
    {
        a+=b;
    }
    return a;
}
int main()
{
    freopen("ariprog.in","r",stdin);
    freopen("ariprog.out","w",stdout);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int num = init();
        int k=0;
        for(int i=0;i<num;i++)
        {
            for(int j=i+1;j<num;j++)
            {
                int b=bisquares[j]-bisquares[i];
                int pre=bisquares[i]-b;
                if(pre>=0&&Set[pre])continue;
                int x=check(bisquares[j],b);
                for(int a=bisquares[i];a+(n-1)*b<=x;a+=b)
                    res[k++]=result(a,b);
            }
        }
        sort(res,res+k);
        if(k==0)
            printf("NONE\n");
        else
            for(int i=0;i<k;i++)
                printf("%d %d\n",res[i].a,res[i].b);
    }
    return 0;
}