https://www.luogu.org/problemnew/show/P1214

题解：枚举前两个数，即可推出公差，从而推出整个数列，判断即可。

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum;
int a[N];
bool vis[N];
char str;
struct node{
    int a,b;
    bool operator <(const node&S)const{
        if(b==S.b)
            return a<S.a;
        return b<S.b;
    }
}e[N];
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=m;i++){
        for(int j=i;j<=m;j++){
            if(!vis[i*i+j*j])a[++cnt]=i*i+j*j;
            vis[a[cnt]]=1;
        }
    }
    sort(a+1,a+cnt+1);
    for(int i=1;i<=cnt-n+1;i++){
        for(int j=i+1;j<=cnt-n+2;j++){
            int d=a[j]-a[i];
            flag=1;
            if(a[i]+(n-1)*d>a[cnt])
                break;
            for(int k=0;k<n;k++){
                temp=a[i]+k*d;
                if(!vis[temp]){
                    flag=0;
                    break;
                }
            }
            if(flag){
                e[++ans].a=a[i];
                e[ans].b=d;
            }
        }
    }
    if(!ans)
        cout<<"NONE"<<endl;
    sort(e+1,e+ans+1);
    for(int i=1;i<=ans;i++){
        printf("%d %d\n",e[i].a,e[i].b);
    }

    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}