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  • 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 nm (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.

    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 = 999999;
     7         for(y = 0; y <= n / 2; y++){
     8             x = n - 2 * y;
     9             if((x + y) % m == 0){
    10                 if((x + y) < min){
    11                     min = x + y;
    12                 }
    13             }
    14         }
    15         if(min != 999999){
    16             printf("%d ", min);
    17         }
    18         else{
    19             printf("-1 ");
    20         }
    21     }
    22     return 0;

    23 } 

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  • 原文地址:https://www.cnblogs.com/angle-qqs/p/4025292.html
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