1031 Hello World for U (20分)

Given any string of N (≥) 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 } 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

题解：
注意n1,n2,n3的大小关系，n1=n2<=n2(n1尽可能的大)
len=strlen(n)+2
n1=len/3;
n2=len/3+len%3;
n3=len/3;

#include<bits/stdc++.h>
using namespace std;
const int maxn=110;

int main(){
    char s[maxn],u[maxn][maxn];
    scanf("%s",s);
    int len=strlen(s)+2;
    int n1,n2,n3;
    n1=len/3;
    n2=len/3+len%3;
    n3=len/3;
    int index=0;
    fill(u[0],u[0]+maxn*maxn,' ');
    for(int i=0;i<n1;i++){
        u[i][0]=s[index++];
    }
    for(int i=1;i<n2-1;i++){
        u[n1-1][i]=s[index++];
    }
    for(int i=0;i<n3;i++){
        u[n3-i-1][n2-1]=s[index++];
    }
    for(int i=0;i<n1;i++){
        for(int j=0;j<n2;j++){
            printf("%c",u[i][j]);
        }
        printf("\n");
    }
    return 0;
}

标签：20,string,int,len,maxn,characters,World,line,1031
来源： https://www.cnblogs.com/dreamzj/p/14350856.html