USACO Section 1.4 - Arithmetic Progressions （暴力）

题意

给出m，要求在[0, m]的范围内，由(0 <= p, q <= m)，p2+q2所组成的集合内，输出长度为n的等差数列。

思路

m最大才250，直接开个vis数组，记录达到的数。

接下来我是把这些数都存在一个数组里，从这个数组里取出来，两两枚举k，然后判断接下来的数是否vis过。

代码

/*
ID: mycodeb1
LANG: C++
TASK: ariprog
*/
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#define LL long long
#define lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 250 + 5;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int vis[MAXN * MAXN * 2];
vector<pii> ans;
vector<int> mp;
int main()
{
    //ROP;
    freopen("ariprog.in", "r", stdin);
    freopen("ariprog.out", "w", stdout);
    int n, m, i, j;
    scanf("%d%d", &n, &m);
    for (i = 0; i <= m; i++)
        for (j = 0; j <= m; j++)
        {
            int a = i * i + j * j;
            if (vis[a]) continue;
            mp.PB(a);
            vis[a] = 1;
        }
    sort(mp.begin(), mp.end());
    for (i = 0; i < mp.size(); i++)
        for (j = i + 1; j < mp.size(); j++)
        {
            int cnt = 2;
            int k = mp[j] - mp[i], last = mp[j];
            if (mp[i] + (n - 1) * k > mp.back()) continue;
            while (cnt < n)
            {
                if (!vis[mp[i] + cnt * k]) break;
                else cnt++;
            }
            if (cnt == n) ans.push_back(MP(k, mp[i]));
        }
    sort(ans.begin(), ans.end());
    if (ans.empty())
    {
        puts("NONE");
        return 0;
    }
    for (vitii it = ans.begin(); it != ans.end(); it++)
        printf("%d %d\n", it->S, it->F);
    return 0;
}