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 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
代码
//题目要求将一行字符串变成U型,n1为左边一列,n2为最下面一行,n3为右边一列
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int Find(char *str,int l)
{
int maxl=0;
//i表示n2
for(int i=3; i<=l; i++)
{
int x=l-i; //x为2*(n1-1)
//n1<=n2,那么n1-1<n2
if(x%2==0&&x/2>=maxl&&x/2<i)
maxl=x/2;
}
return maxl;
}
int main()
{
char str[100];
while(~scanf("%s",str))
{
int l=strlen(str);
int n=Find(str,l);
//printf("%d\n",n);
int c=0;
while(c<n)
{
printf("%c",str[c]);
for(int i=0; i<l-2*n-2; i++)
printf(" ");
printf("%c\n",str[l-1-c]);
c+=1;
}
for(int i=c; i<l-c; i++)
printf("%c",str[i]);
}
return 0;
}