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: 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.
输入描述:
There are multiple test cases.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.
输出描述:
For each test case, print the input string in the shape of U as specified in the description.
示例1
输入
helloworld!
www.nowcoder.com
输出
h !
e d
l l
lowor
w m
w o
w c
. .
n r
owcode
解答
#include<iostream>
#include<string>
#include<stdio.h>
using namespace std;
char str[81];
int n,n1,n2;
// 穷举法,解方程:
void calculateN(){
n2 = 3;
for (int i = 0; i < n; i++)
{
if (n2==n) break;
n1=0; // n1 position;
for (int j = 0; j < n2; j++)
{
n1++;
if (n1*2+n2-2==n)
{
return;
}
}
n2++;
}
}
// string 的知识:
int main(){
string temp;
while(getline(cin,temp)){
// cout<<temp;
n = temp.size();
calculateN(); // 计算出 n1、n2
char str[n1][n2];//
// cout<<n<<n1<<n2;
// fill the matrix str
for (int i = 0; i < n1; i++)
{
for (int j = 0; j < n2; j++)
{
if (i!=n1-1)
{
if(j==0){
str[i][j] = temp.at(0);
temp.erase(0,1); // 从0位置开始,包括这个位置的 n-1的字符
}
else if(j==n2-1){
str[i][j] = temp.at(temp.size()-1);
temp.erase(temp.size()-1,1);
}else
{
str[i][j] = ' ';
}
}else{
str[i][j] = temp.at(0);
temp.erase(0,1);
}
}
}
// output
for(int i=0;i<n1;i++){
for(int j=0;j<n2;j++)
cout<<str[i][j];
cout<<endl;
}
}
return 0;
}
总结:
- String 的知识:
- 矩阵和二维数组直接的关系
第一个维度对应行
第二个维度对应列