1031. Hello World for U (20)
2016-02-22 16:23
597 查看
Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as:
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line
with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3
- 2 = N.
Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
Output Specification:
For each test case, print the input string in the shape of U as specified in the description.
Sample Input:
Sample Output:
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
string s;
cin >> s;
int n = (int)s.size() + 2;
int n1, n2, n3;
int r = n % 3;
if(r == 0){
n1 = n2 = n3 = n / 3;
}else{
n2 = (n - r) / 3 + r;
n1 = n3 = (n - r) / 3;
}
int i = 0, j = (int)s.size() - 1;
for(; i < n1-1; ++i, --j){
printf("%c", s[i]);
for(int k = 0; k < n2-2; ++k)
printf(" ");
printf("%c\n", s[j]);
}
for(; i < n1+n2-1; ++i){
printf("%c", s[i]);
}
return 0;
}
h d e l l r lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line
with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3
- 2 = N.
Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
Output Specification:
For each test case, print the input string in the shape of U as specified in the description.
Sample Input:
helloworld!
Sample Output:
h ! e d l l lowor
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
string s;
cin >> s;
int n = (int)s.size() + 2;
int n1, n2, n3;
int r = n % 3;
if(r == 0){
n1 = n2 = n3 = n / 3;
}else{
n2 = (n - r) / 3 + r;
n1 = n3 = (n - r) / 3;
}
int i = 0, j = (int)s.size() - 1;
for(; i < n1-1; ++i, --j){
printf("%c", s[i]);
for(int k = 0; k < n2-2; ++k)
printf(" ");
printf("%c\n", s[j]);
}
for(; i < n1+n2-1; ++i){
printf("%c", s[i]);
}
return 0;
}
相关文章推荐
- 使用C++实现JNI接口需要注意的事项
- 关于指针的一些事情
- c++ primer 第五版 笔记前言
- share_ptr的几个注意点
- Lua中调用C++函数示例
- Lua教程(一):在C++中嵌入Lua脚本
- Lua教程(二):C++和Lua相互传递数据示例
- C++联合体转换成C#结构的实现方法
- C++高级程序员成长之路
- C++编写简单的打靶游戏
- C++ 自定义控件的移植问题
- C++变位词问题分析
- C/C++数据对齐详细解析
- C++基于栈实现铁轨问题
- C++中引用的使用总结
- 使用Lua来扩展C++程序的方法
- C++中调用Lua函数实例
- Lua和C++的通信流程代码实例
- C与C++之间相互调用实例方法讲解
- 解析C++中派生的概念以及派生类成员的访问属性