A1031 Hello World for U
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
idea
- 限制条件为:
n1 = n3 <=n2
其中,n1 + n2 + n3 = n + 2(n为所给字符串长度
加2是因为两个绿色的角角多算了一遍
==》则n1 = n2 = (n + 2) / 3 , n2 = n + 2 - 2 * n1
solution1
#include <stdio.h>
#include <string.h>
int main(){
char c[100];
int n, n1, n2, k = 0;
scanf("%s", c);
n = strlen(c);
n1 = (n + 2) / 3;
n2 = n - 2 * n1 + 2;//两个底角重复了一遍
char a[n1][n2];
for(int i = 0; i < n1; i++)
a[i][0] = c[k++];
for(int i = 0; i < n1; i++)
for(int j = 1; j < n2 - 1; j++) {
if(i != n1 - 1)
a[i][j] = ' ';
else
a[i][j] = c[k++];
}
for(int i = n1 - 1; i >= 0; i--)
a[i][n2 - 1] = c[k++];
for(int i = 0; i < n1; i++){
for(int j = 0; j < n2; j++)
printf("%c", a[i][j]);
if(i != n1 - 1)
printf("\n");
}
return 0;
}
solution(优化版
#include <stdio.h>
#include <string.h>
int main(){
char s[81];
scanf("%s", s);
int n = strlen(s), n1 = (n + 2) / 3, n2 = n + 2 - 2 * n1,num = 0;
char a[n1][n2];
memset(a, ' ', sizeof(a));
for(int i = 0; i < n1; i++)
a[i][0] = s[num++];
for(int i = 1; i < n2; i++)
a[n1 - 1][i] = s[num++];
for(int i = n1 - 2; i >= 0; i--)
a[i][n2 - 1] = s[num++];
for(int i = 0; i < n1; i++){
for(int j = 0 ; j < n2; j++){
printf("%c", a[i][j]);
}
printf("\n");
}
return 0;
}