PAT 1031 Hello World for U

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 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

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;
}