zoukankan      html  css  js  c++  java
  • 小虫爬井

    Problem Description

    An inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minute before climbing again. During the rest, it slips down d inches. The process of climbing and resting then repeats. How long before the worm climbs out of the well? We'll always count a portion of a minute as a whole minute and if the worm just reaches the top of the well at the end of its climbing, we'll assume the worm makes it out.

    Input

    There will be multiple problem instances. Each line will contain 3 positive integers n, u and d. These give the values mentioned in the paragraph above. Furthermore, you may assume d < u and n < 100. A value of n = 0 indicates end of output.

    Output

    Each input instance should generate a single integer on a line, indicating the number of minutes it takes for the worm to climb out of the well.

    Sample Input

    10 2 1 20 3 1 0 0 0

    Sample Output

    17 19

    #include<stdio.h>
    int main()
    {
        int n, u, d, t;
        int i = 0, j = 0, k;
        while( scanf( "%d%d%d" , &n, &u, &d ) == 3)
        {
              t = 0;
              i = 0;
              if( n == 0 && u == 0 && d == 0 )
                       break;
               while( 1 )
               {
                    i += u ;
                    if( i >= n )
                    {
                      t++;
                      break;
                      }
                    i -= d;
                    t += 2;
        
               }
           printf( "%d\n", t );
                
        }
    }

  • 相关阅读:
    遍历Map集合的方法
    简易循环手动增加表的数据
    初识Mysql,mysql的介绍和一些SQL语句
    SQL语句
    STS中applicationContext.xml配置文件
    STS中poem.xml配置文件
    MybatisMapper 动态映射(增删改查)
    MybatisMapper 映射框架(增删改查 原始模式)
    Spring 框架下 (增 删 改 )基本操作
    Spring 框架下 事务的配置(复杂)
  • 原文地址:https://www.cnblogs.com/zsj576637357/p/2250533.html
Copyright © 2011-2022 走看看