代码练习系列:问题 B Hello World for U
2017-11-02 19:29
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题目描述
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
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.
输出
For each test case, print the input string in the shape of U as specified in the description.
样例输入
helloworld!
样例输出
h !
e d
l l
lowor
提示
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
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.
输出
For each test case, print the input string in the shape of U as specified in the description.
样例输入
helloworld!
样例输出
h !
e d
l l
lowor
提示
#include <stdio.h> #include <string.h> int main() { char s[80]; gets(s); int num = strlen(s); int n1,n2,mid; n1 = n2 = (num + 2) / 3; mid = num - 2 * n1; for(int i = 0;i < n1;i++){ printf("%c",s[i]); if(i == n1 - 1){ for(int k = 0;k < mid;k++){ printf("%c",s[n1 + k]); } }else{ for(int j = 0;j < mid;j++){ printf(" "); } } printf("%c\n",s[num - 1 - i]); } return 0; }
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