『算法-ACM 竞赛-数学-数论』 HDU - 2854 Central Meridian Number （暴力打表）

A Central Meridian (ACM) Number N is a positive integer satisfies that given two positive integers A and B, and among A, B and N, we have
N | ((A^2)*B+1) Then N | (A^2+B)
Now, here is a number x, you need to tell me if it is ACM number or not.

Input

The first line there is a number T (0<T<5000), denoting the test case number.
The following T lines for each line there is a positive number N (0<N<5000) you need to judge.

Output

For each case, output “YES” if the given number is Kitty Number, “NO” if it is not.

Sample Input

2
3
7
Sample Output

YES
NO
Hint
X | Y means X is a factor of Y, for example 3 | 9;
X^2 means X multiplies itself, for example 3^2 = 9;
XY means X multiplies Y, for example 33 = 9.

题意：

给你一个数，如果能找出两个数 a,b 使得这三个数满足式子 1，但不满足式子 2，那么这个数 n 就不是符合要求的数，输出 NO

思路：
实在算不粗来了，把我写的代码打了个表，然后，发现最大是 240，然后其他，就没其他了。

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
int mp[10000000];
int main()
{
    mp[1] = 1;
    mp[2] = 1;
    mp[3] = 1;
    mp[4] = 1;
    mp[5] = 1;
    mp[6] = 1;
    mp[8] = 1;
    mp[10] = 1;
    mp[12] = 1;
    mp[15] = 1;
    mp[16] = 1;
    mp[20] = 1;
    mp[24] = 1;
    mp[30] = 1;
    mp[40] = 1;
    mp[48] = 1;
    mp[60] = 1;
    mp[80] = 1;
    mp[120] = 1;
    mp[240] = 1;
    int T, a;
    cin >> T;
    while (T--)
    {
        scanf("%d", &a);
        if (mp[a] == 1)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

达标代码

#include <iostream>
#include <cstdio>
using namespace std;
int mp[10000000];

int ok(int x)
{
    for (int i = 1; i <= 1000; i++)
    {
        for (int j = 1; j <= 1000; j++)
        {
            if ((i * i * j + 1) % x == 0 && ((i * i + j) % x != 0))
                return 0;
        }
    }
    return 1;
}
int main()
{
    for (int i = 1; i <= 1000; i++) //打表部分
    {
        if (ok(i))
        {
            printf("mp[%d]=1; \n", i);
        }
    }
}