题目描述

由题可知
$n = 2*n1 + n2$
$n1 = n3$
$n2 >= n1$

分析
$(n + 2) \% 3 = 0， n2 = n1$
$(n + 2) \% 3 = 1， n2 = n1 + 1$
$(n + 2) \% 3 = 2， n2 = n1 + 2$

所以可得
$n1 = (n + 2) / 3$
$n2 = n1 + (n + 2) \% 3$

C++ 代码

#include <iostream>
#include <string>
using namespace std;

const int MAX_N = 81;
char g[MAX_N][MAX_N];

int main() {
    string str;
    cin >> str;
    int n = str.length();
    int n1 = (n + 2) / 3;
    int n3 = n1;
    int n2 = n1 + (n + 2) % 3;

    int cnt = 0;
    for (int i = 0; i < n1; i++) g[i][0] = str[cnt++];
    for (int i = 1; i < n2; i++) g[n1 - 1][i] = str[cnt++];
    for (int i = n3 - 2; i >= 0; i--) g[i][n2 - 1] = str[cnt++];

    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
            if (g[i][j]) printf("%c", g[i][j]);
            else printf(" ");           
        }
        printf("\n");
    }
    return 0;
}