Dreamoon and Stairs

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本文最后更新于 2023-11-17,文中内容可能已过时。

题目链接

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?

1 Input

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

2 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.

3 Examples

3.1 input1

10 2

3.2 output1

6

3.3 input2

3 5

3.4 output2

-1

3.5 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.

有一个 n 级台阶,每次可以走一级或两级,问最少的步数是多少,且步数必须是 m 的倍数。
找一下数学公式就好了。
具体看代码。

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#include<bits/stdc++.h>
using namespace std;

int main(){
    int x,n,m;
    cin>>n>>m;
    if(n<m){
        cout<<-1<<endl;
        return 0;
    }
    if(n==m){
        cout<<n<<endl;
        return 0;
    }
    if(n%2==0){
        x=n/2%m;
        if(x==0) cout<<n/2<<endl;
        else cout<<n/2+m-x<<endl;
    }else if(n%2!=0){
        x=(n/2+1)%m;
        if(x==0) cout<<n/2+1<<endl;
        else cout<<(n/2+1)+m-x<<endl;
    }
    return 0;
}

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