A. Dreamoon and Stairs(Codeforces Round #272)
2016-04-04 15:09
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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 steps 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
output
input
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.
找到范围,推断一下全部这个范围的点即可了。
代码:
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 steps 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
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.
找到范围,推断一下全部这个范围的点即可了。
代码:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int n,m; int l; scanf("%d%d",&n,&m); if(n%2==0) { l=n/2; } else l=n/2+1; int ans=-1; for(int i=l;i<=n;i++) if(i%m==0) { ans=i; break; } printf("%d\n",ans); return 0; }
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