字符串構造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.

輸入描述:
There are multiple test cases.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.

輸出描述:
For each test case, print the input string in the shape of U as specified in the description.
示例1
輸入

helloworld!
www.nowcoder.com

輸出

h   !
e   d
l   l
lowor
w    m
w    o
w    c
.    .
n    r
owcode

解答

#include<iostream>
#include<string>
#include<stdio.h>

using namespace std;

char str[81];
int n,n1,n2;

// 窮舉法，解方程：
void calculateN(){
    n2 = 3;
        for (int i = 0; i < n; i++)
        {
            if (n2==n) break;
            n1=0; // n1 position;
            for (int j = 0; j < n2; j++)
            {
                n1++; 
                if (n1*2+n2-2==n)
                {
                    return;
                }
                
            }
            n2++;
        }
}
 // string 的知識：

int main(){
    string temp;
    while(getline(cin,temp)){
        // cout<<temp;
        n = temp.size();
        calculateN(); // 計算出 n1、n2
        char str[n1][n2];//
       // cout<<n<<n1<<n2;

        // fill the matrix str
        for (int i = 0; i < n1; i++)
        {
            for (int j = 0; j < n2; j++)
            {  
                if (i!=n1-1)
                {
                    if(j==0){
                        str[i][j] = temp.at(0);
                        temp.erase(0,1); // 從0位置開始，包括這個位置的 n-1的字符
                    }
                    else if(j==n2-1){
                        str[i][j] = temp.at(temp.size()-1);
                        temp.erase(temp.size()-1,1); 
                    }else
                    {
                        str[i][j] = ' ';
                    }
                }else{
                    str[i][j] = temp.at(0);
                    temp.erase(0,1);
                }
                
            }
        }
        // output
        for(int i=0;i<n1;i++){
            for(int j=0;j<n2;j++)
                cout<<str[i][j];
            cout<<endl;
        }
    }
    return 0;
}

總結：

1. String 的知識：
2. 矩陣和二維數組直接的關係
   第一個維度對應行
   第二個維度對應列

string的知識