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

A. Dreamoon and Stairs

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

6

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
");
#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;
}