luogu P1214 等差数列 枚举

//打出表可以看到，集合内有大约cnt=120000个数。做法是枚举等差数列前两个数，然后再用一个N去判断序列后面每一位是否存在。复杂度高达cnt^2*n，加了玄学优化可以过，应该不是正解。 
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector <int> vec;
int n,m;
int vis[130000];
struct dat
{
    int a,b;
    dat(int _a = 0,int _b = 0):a(_a),b(_b){}
    friend bool operator < (dat x,dat y)
    {
        if (x.b != y.b)
            return x.b < y.b;
        return x.a < y.a; 
    }
};
vector <dat> ans;
int main()
{
    scanf("%d%d",&n,&m);
    for (int i = 0;i <= m;i++)
        for (int j = i;j <= m;j++)
            vis[i * i + j * j] = true;
    for (int i = 0;i <= 125000;i++)
        if (vis[i] == true)
            vec.push_back(i);
    for (int i = 0;i < vec.size();i++)
        for (int j = i + 1;j < vec.size();j++)
        {
            int a = vec[i],b = vec[j] - vec[i];
            //玄学剪枝 
            if (a + (n - 1) * b > vec[vec.size() - 1])
                break;
            int cnt = 1;
            while (vis[a + cnt * b] == true && cnt < n)
                cnt++;
            if (cnt == n)
                ans.push_back(dat(a,b));
        }
    if (ans.size() == 0)
    {
        printf("NONE
");
        return 0;
    }
    sort(ans.begin(),ans.end());
    for (int i = 0;i < ans.size();i++)
        printf("%d %d
",ans[i].a,ans[i].b);
    return 0;
}