UVA_11636_HelloWorld!
2015-09-02 21:21
357 查看
11636 - Hello World!
Time limit: 1.000 secondsWhen you rst made the computer to print the sentence \Hello World!", you felt so happy, not
knowing how complex and interesting the world of programming and algorithm will turn out to be.
Then you did not know anything about loops, so to print 7 lines of \Hello World!", you just had to
copy and paste some lines. If you were intelligent enough, you could make a code that prints \Hello
World!" 7 times, using just 3 paste commands. Note that we are not interested about the number
of copy commands required. A simple program that prints \Hello World!" is shown in Figure 1. By
copying the single print statement and pasting it we get a program that prints two \Hello World!"
lines. Then copying these two print statements and pasting them, we get a program that prints four
\Hello World!" lines. Then copying three of these four statements and pasting them we can get a
program that prints seven \Hello World!" lines (Figure 4). So three pastes commands are needed in
total and Of course you are not allowed to delete any line after pasting. Given the number of \Hello
World!" lines you need to print, you will have to nd out the minimum number of pastes required to
make that program from the origin program shown in Figure 1.
Input
The input le can contain up to 2000 lines of inputs. Each line contains an integer N (0 < N < 10001)
that denotes the number of \Hello World!" lines are required to be printed.
Input is terminated by a line containing a negative integer.
Output
For each line of input except the last one, produce one line of output of the form `Case X: Y ' where
X is the serial of output and Y denotes the minimum number of paste commands required to make a
program that prints N lines of \Hello World!".
Sample Input
2
10
-1
Sample Output
Case 1: 1
Case 2: 4
今天在UVA注册了账号,用这个helloworld记录下吧
这个题目说的可以复制一部分,因此就找大于它的最近的二进制值就可以了
#include <iostream> #include <stdio.h> using namespace std; int er[20]={1}; int main() { int ca=1; for(int i=1;er[i-1]<10005;i++) er[i]=er[i-1]*2; int n; while(scanf("%d",&n)!=EOF) { if(n<0) break; int p=0; while(n>er[p]) p++; printf("Case %d: %d\n",ca++,p); } return 0; }
相关文章推荐
- 解决ViewPager.setCurrentItem不能实现平滑移动
- ZOJ 2872 Binary Partitions
- 树状数组 离散化 求逆序数POJ 2299Ultra-QuickSort解题报告
- 求N!的末尾0的个数
- C++操作符的优先级
- APP网络查询数据,多客户端分页缓存
- 红色警戒2修改器原理百科(十)
- iptables 执行清除命令 iptables -F 要非常小心的
- iOS开发 main.m
- hdu 1241 深搜水题
- bzoj1028 [JSOI2007]麻将
- C++模板元编程 - 2 模仿haskell的列表以及相关操作
- 关于sed -i 修改selinux 的软链接文件的问题
- 排序代码
- CERC-2014 K题 - The Imp (博弈DP)
- C++--------------------------------------------运算符重载
- Fragment的用法(建议用support.v4的包,所以导包时认真注意,用getsupport的那个方法)
- C语言介绍
- 成功的背后!(给所有IT人)
- PHP的SOAP工具包--nusoap