CodeForce-476A-Dreamoon and stairs

原题：
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 moves 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.
Examples
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：
小明要上n阶台阶，每一次跨越只能上一步或者两步，小明想以m的最小跨越数上完所有的台阶，问最少需要跨越几次？若是无法满足题意则输出-1.
题解：
一道水题，如果m大于台阶数，则一定不会以m的倍数次跨越次数上完台阶，此时输出-1.
如果m小于台阶数，由于每次只能上一阶或者两阶，所以只需计算出上完所有台阶所需最少步数mini=n/2（向上取整）。如果不是m的倍数，则只需将之前的两步分成两个一步，就可以凑成m倍。所以当m小于台阶数一定可以满足题意的上完台阶。（具体请理解代码）
附上AC代码：

#include <iostream>
#include <cmath>
using namespace std;
double n,m,ans;
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        int maxi=n;//记录最大台阶数
        if(m>maxi)//如果m大于最大台阶数，则一定不能以m的倍数上完台阶
        {
            cout<<-1<<endl;
            continue;
        }
        int mini=ceil(n/2);//记录上完台阶的最小跨越数
        mini=ceil(mini/m);//记录下mini是m的几倍，向上取整
        int i=m*mini;
        cout<<i<<endl;
    }
    return 0;
}

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CodeForce-476A-Dreamoon and stairs

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