数学--数论-- HDU -- 2854 Central Meridian Number （暴力打表）
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);
}
}
}