1031 Hello World for U
1031 Hello World for U
0、题目
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.
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
1、大致题意
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。并满足一下要求:
- n1 == n3
- n2 >= n1
- n1为在满足上述条件的情况下的最大值
2、基本思路
分三种情况讨论:
- 如果 n % 3 = = 0 n \% 3 == 0 n%3==0 ,直接 n 1 = n 2 = n 3 n1 = n2 = n3 n1=n2=n3
- 如果 n % 3 = = 1 n \% 3 == 1 n%3==1,因为 n 2 n2 n2 要比 n 1 n1 n1 大,所以把多出来的那1个给 n 2 n2 n2
- 如果 n % 3 = = 2 n \% 3 == 2 n%3==2,就把多出来的那2个给 n 2 n2 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。
3、AC代码
#include <iostream>
#include <string.h>
using namespace std;
int main() {
char c[81], u[30][30];
memset(u, ' ', sizeof(u));
scanf("%s", c);
int n = strlen(c) + 2;
int n1 = n / 3, n2 = n / 3 + n % 3, 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;
}
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