zoukankan      html  css  js  c++  java
  • uva-----(11384)Help is needed for Dexter

    Problem H

    Help is needed for Dexter

    Time Limit: 3 Second

     

    Dexter is tired of Dee Dee. So he decided to keep Dee Dee busy in a game. The game he planned for her is quite easy to play but not easy to win at least not for Dee Dee. But Dexter does not have time to spend on this silly task, so he wants your help.

    There will be a button, when it will be pushed a random number N will be chosen by computer. Then on screen there will be numbers from 1 to N. Dee Dee can choose any number of numbers from the numbers on the screen, and then she will command computer to subtract a positive number chosen by her (not necessarily on screen) from the selected numbers. Her objective will be to make all the numbers 0.

    For example if N = 3, then on screen there will be 3 numbers on screen: 1, 2, 3. Say she now selects 1 and 2. Commands to subtract 1, then the numbers on the screen will be: 0, 1, 3. Then she selects 1 and 3 and commands to subtract 1. Now the numbers are 0, 0, 2. Now she subtracts 2 from 2 and all the numbers become 0.

    Dexter is not so dumb to understand that this can be done very easily, so to make a twist he will give a limit L for each N and surely L will be as minimum as possible so that it is still possible to win within L moves. But Dexter does not have time to think how to determine L for each N, so he asks you to write a code which will take N as input and give L as output.

    Input and Output:

    Input consists of several lines each with N such that 1 ≤ N ≤ 1,000,000,000. Input will be terminated by end of file. For each N output L in separate lines.

    SAMPLE INPUT

    OUTPUT FOR SAMPLE INPUT

    1

    2

    3

    1

    2

    2

    Problemsetter: Md. Mahbubul Hasan

    代码:

     1 #include<stdio.h>
     2 int work(int n)
     3 {
     4   int ans=0;
     5   while(n>0){
     6    ans++;
     7    n>>=1;
     8  }
     9   return ans;
    10 }
    11 int main()
    12 {
    13     int n;
    14     while(scanf("%d",&n)!=EOF)
    15       printf("%d
    ",work(n));
    16     return 0;
    17 }
    View Code
  • 相关阅读:
    helix matrix 螺旋矩阵
    利用基数排序生成100个1000以内顺序排列的随机数
    第三次随笔
    gSoap: How to add info to SOAP Header using gSOAP(转)
    sizeof empty class object
    C++基础回顾字符串地址比较
    (转)RVA相对虚拟地址解释
    团队建设历史经验回顾
    C++全局变量的生老病死
    C++基础回顾强制类型转换
  • 原文地址:https://www.cnblogs.com/gongxijun/p/3891200.html
Copyright © 2011-2022 走看看