A. Dreamoon and Stairs（Codeforces Round #272)

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最新推荐文章于 2024-04-18 15:20:21 发布

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本文介绍了一个有趣的算法问题——如何让角色Dreamoon以步数为特定整数倍的方式爬上n阶楼梯，并给出了高效的解决方案及代码实现。

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Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer m.

What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?

Input

The single line contains two space separated integers n, m (0 < n ≤ 10000, 1 < m ≤ 10).

Output

Print a single integer — the minimal number of moves being a multiple of m. If there is no way he can climb satisfying condition print - 1instead.

Sample test(s)

input
10 2

output
6

input
3 5

output
-1

Note

For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.

For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.

找到范围，判断一下所有这个范围的点就行了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int n,m;
    int l;
    scanf("%d%d",&n,&m);
    if(n%2==0)
    {
        l=n/2;
    }
    else
    l=n/2+1;
    int ans=-1;
    for(int i=l;i<=n;i++)
    if(i%m==0)
    {
        ans=i;
        break;
    }
    printf("%d\n",ans);
    return 0;
}