Pat(Advanced Level)Practice--1031(Hello World for U)
Pat 1031コード
タイトルの説明:
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:
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:
Sample Output:
ACコード:
タイトルの説明:
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
ACコード:
#include<cstdio>
#include<cstring>
#define MAX 100
using namespace std;
int main(int argc,char *argv[])
{
int len;
char str[MAX];
int i,j;
int n1,n2,n3;
scanf("%s",str);
len=strlen(str);
n1=n3=(len+2)/3;
n2=(len+2)%3+n1;
for(i=0;i<n1;i++)
{
printf("%c",str[i]);
if(i<n1-1)
{
for(j=3;j<=n2;j++)
printf(" ");
}
else
{
for(j=3;j<=n2;j++)
printf("%c",str[i+j-2]);
}
printf("%c
",str[len-i-1]);
}
return 0;
}