Codeforces Round #272 (Div. 2)A. Dreamoon and Stairs 水题

最新推荐文章于 2025-08-21 10:38:03 发布

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最新推荐文章于 2025-08-21 10:38:03 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zxhl/p/4930357.html

本文针对一个经典的算法问题——爬楼梯，给出了详细的解析过程。问题要求找到达到第n级阶梯所需的步数，且总步数必须为给定整数m的倍数。文章通过示例解释了如何使用算法来寻找最小的满足条件的步数。

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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 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 - 1instead.

Sample test(s)

input

10 2

output

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：给你n,m，每次可以走1或2，问你走到n，并且走路的次数是m的倍数，问走几次

题解：先尽量分为2多,1少，再将2分出1,1，，即是最小m的倍数

///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define inf 100000000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************
#define maxn 10000+5

int main(){
    int n=read();
    int ans=0;
    int m=read();
    if(n<m){cout<<-1<<endl;return 0;}
    int tmp=n/2; ans=n%2+tmp;
    if(ans%m==0){
        cout<<ans<<endl;
    }
    else {
        ans=(ans+m-ans%m);
        cout<<ans<<endl;
    }

  return 0;
}