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

http://codeforces.com/contest/476/problem/A

A. Dreamoon and Stairs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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.

解题思路：一个N阶的楼梯，每次能走1，或者2阶，求最小的爬完楼梯的次数，且要满足次数是m的倍数

贪心，枚举最少走2阶且满足条件的次数

 1 #include <stdio.h>

 2 

 3 int main(){

 4     int x, y, n, m, min;

 5     while(scanf("%d %d", &n, &m) != EOF){

 6         min = ;

 7         for(y = ; y <= n / ; y++){

 8             x = n -  * y;

 9             if((x + y) % m == ){

                 if((x + y) < min){

                     min = x + y;

                 }

             }

         }

         if(min != ){

             printf("%d\n", min);

         }

         else{

             printf("-1\n");

         }

     }

     return ;

23 }

巴特西

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

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