AtCoder Beginner Contest 182B

B - Almost GCD

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $200$ points

Problem Statement

Given is an integer sequence $A$: ${\mathrm{A1,A2,A3,\dots ,AN}}^{}$.
Let the GCD-ness of a positive integer $k$ be the number of elements among ${\mathrm{A1,A2,A3,\dots ,AN}}^{}$ that are divisible by $k$.
Among the integers greater than or equal to $2$, find the integer with the greatest GCD-ness. If there are multiple such integers, you may print any of them.

Constraints

• $\mathrm{1\le N\le 100\le N\le 100}$
• $\mathrm{2\le Ai\le 1000}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$

${\mathrm{A1A2A3\dots AN}}^{}$

Output

Print an integer with the greatest GCD-ness among the integers greater than or equal to $22$. If there are multiple such integers, any of them will be accepted.

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Sample Input 1 

3

3 12 7

Sample Output 1 

3

Among $3$, $12$, and $7$, two of them - $3$ and $12$ - are divisible by $3$, so the GCD-ness of $3$ is $2$.

No integer greater than or equal to $2$ has greater GCD-ness, so $3$ is a correct answer.

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Sample Input 2 

5

8 9 18 90 72

Sample Output 2 

9

In this case, the GCD-ness of $9$ is $4$

$2$ and $3$ also have the GCD-ness of $4$, so you may also print $2$ or $3$.

---

Sample Input 3 

5

1000 1000 1000 1000 1000

Sample Output 3

1000

解题思路：拿到手，第一眼(没看题)看了输入的形式加上有个GCD我还以为要靠贝祖定理了呢，然后仔细分析是求一个K，K满足两个条件

①：K是输入的数组的元素的公因数，且从2开始。

②：k的倍数的数最多(当然这样的话就会有很多情况，随便输出一种情况就行)

 再一看数据，这么小直接二重循环暴力出来啦，外层循环从2开始，一直到1000，内次循环遍历数组，每次内存循环记录满足倍数条件的个数

Code:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
 
int a[1005];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d",&a[i]);
    }
    int max_ = 0,max_key = a[1],loc;
    for(int i = 2; i <= 1000; ++i) {
        loc = 0;//表示的是每次循环i的倍数的个数(当然是对a来说的)
        for(int j = 1; j <= n; ++ j) {
            if(a[j] % i == 0)
            loc++;
        }
        if(loc > max_) {//如果i的倍数比记录的大，就更新
            max_ = loc;
            max_key = i;
        }
    }
    printf("%d\n",max_key);
    return 0;
}

 

 开始以为k只能是a数组的元素，wa了很多发T^T