UVA11636-Hello World!
Time limit: 1.000 seconds
When 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
题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2683
题目大意：给定n行，求要复制的次数！
分析：这样题意可能比较难理解，(我英文不太好)我是看样例猜答案的，输入的数很有规律，比如样例1，输入2，输出为1,我就这样想，2/2=1，很有道理，再看样例2，输入10，输出按照前面的想法应该是3才对啊，10/2=5,5/2=2,2/2=1->3，稍微拿脑袋想一想就知道不对了，如果向上取整的话是不是就是正解呢，样例1->(2+1)/2=1,样例2->(10+1)/2=5,(5+1)/2=3,(3+1)/2=2,(2+1)/2=1,刚好就等于4，输入小于0的数进行特判即可，于是我就进行了大胆的猜测题意：输入的数每次进行向上取整，求这个数需要经过多少次才能变成1！
下面给出AC代码：
1 #include <bits/stdc++.h>
2 using namespace std;
3 int n;
4 int main()
5 {
6 int p=1;
7 while(scanf("%d",&n)!=EOF)
8 {
9 if(n<0)
10 break;
11 int ans=0;
12 while(n>1)
13 {
14 ans++;
15 n=(n+1)/2;
16 }
17 printf("Case %d: %d\n",p++,ans);
18 }
19 return 0;
20 }