Codeforces Round #272 (Div. 2) A-优快云博客

本文深入解析了“Dreamoon and Stairs”问题，即在给定楼梯数量和最小步骤倍数条件下，求解最小步骤数。通过详细分析和代码实现，展示了从全1分解到全2分解的过程，并提供了验证方法。此题旨在提升读者在分解问题、枚举验证及优化算法设计方面的能力。

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题目：

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 - 1 instead.

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.

题意分析：

水题一枚，给出一个N。然后分解N，只能分解为1和2。考虑全部分解为1的情况，则最多需要N步，全部分解为2的情况，如果为奇数N/2+1 偶数N/2.然后枚举，验证就行了。

代码：

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>


using namespace std;


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