1031 Hello World for U (20 分)

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 n​1​​characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3≤n​2​​≤N } with n​1​​+n​2​​+n​3​​−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

 

题目大意：用所给字符串按U型输出。n1和n3是左右两条竖线从上到下的字符个数，n2是底部横线从左到右的字符个数。要求： 1. n1 == n3 2. n2 >= n1 3. n1为在满足上述条件的情况下的最大值

分析：假设n = 字符串长度 + 2，因为2 * n1 + n2 = n，且要保证n2 >= n1, n1尽可能地大，分类讨论： 1. 如果n % 3 == 0，n正好被3整除，直接n1 == n2 == n3; 2. 如果n % 3 == 1，因为n2要比n1大，所以把多出来的那1个给n2 3. 如果n % 3 == 2, 就把多出来的那2个给n2 所以得到公式：n1 = n / 3，n2 = n / 3 + n % 3

把它们存储到二维字符数组中，一开始初始化字符数组为空格，然后按照u型填充进去，最后输出这个数组u。  

#include<iostream> using namespace std; int main(void){ std::ios::sync_with_stdio( false); string s; getline( cin ,s ); int n1=0; for( int i =1; i < s.length(); i++ ){ if( i <= ( s.length() - i*2 + 2 ) ){ n1=i; } else break; } int n2 = s.length() - n1*2; int l = 0 ,h = s.length() - 1; for( int i =1;i < n1 ; i++ ){ cout<<s[l++]; for( int j =1 ; j<=n2;j++) cout<<" "; cout<<s[h--]<<endl; } for( int j =1 ;j<=n2+2;j++) cout<<s[l++]; } #include <iostream> #include <string.h> using namespace std; int main( void ) { char c[81], u[30][30]; memset(u, ' ', sizeof(u)); scanf("%s", c); int n = strlen(c) + 2; int n1 = n / 3; int n2 = n1 + n % 3; int index = 0; for(int i = 0; i < n1; i++) u[i][0] = c[index++]; for(int i = 1; i <= n2 - 2; i++) u[n1-1][i] = c[index++]; for(int i = n1 - 1; i >= 0; i--) u[i][n2-1] = c[index++]; for(int i = 0; i < n1; i++) { for(int j = 0; j < n2; j++) printf("%c", u[i][j]); printf("\n"); } return 0; }