PAT-A 1031. Hello World for U (20)

传送门

https://www.patest.cn/contests/pat-a-practise/1031

题目

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

分析

首先这道题的题意不用读,光看图就看懂了,重要的就是计算行数和列数。

只要按照题目给定的条件算出来行数和列数即可,题目保证n1是最大值,所以从n2为最小值3开始计算,直到满足条件:

1.n1 + n2为偶数
2.n1 <= n2

时停止,然后根据n1和n2输出图形,方法见代码。

源代码

#include <iostream>
#include <string>

using namespace std;

int main(){
    int n1, n2;
    string s;
    cin >> s;
    int n = s.size();
    // 先求出行数n1和列数n2
    for (n2 = 3; n2 <= n; ++n2){
        int tmp = n + 2 - n2; // tmp = n1 + n3 = 2 * n1;
        if (tmp % 2 != 0){
            continue;
        }
        n1 = tmp / 2;
        if (n1 <= n2){
            break;
        }
    }
    int blank = n2 - 2;
    int row = n1 - 1;
    for (int i = 0; i < n1; ++i){ // 行数
        if (i < row){ // 前三行
            printf("%c", s[i]);
            cout << string(blank, ' ');
            printf("%c\n", s[n - 1 - i]);
        }
        else{
            for (int j = 0; j < n2; ++j){ // 最后一行
                printf("%c", s[i + j]);
            }
        }
    }
    return 0;
}

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