解法一：数学

根据数学性质算出 n1 ， n2 ， n3 的大小，然后按格式输出。

#include <bits/stdc++.h>
using namespace std;
int main() {
    int N, n1, n2, n3;
    string str;
    cin >> str;
    N = str.length();
    switch (N % 3) {
        case 0:
            n1 = n3 = N / 3;
            n2 = N / 3 + 2;
            break;
        case 1:
            n1 = n2 = n3 = (N + 2) / 3;
            break;
        case 2:
            n1 = n3 = (N + 1) / 3;
            n2 = n1 + 1;
    }
    int head = 0, tail = N - 1;
    for (int i = 0; i < n1 - 1; ++i, ++head, --tail) {
        cout << str[head];
        for (int j = 0; j < n2 - 2; ++j) {
            cout << " ";
        }
        cout << str[tail] << '\n';
    }
    for (int i = head; i <= tail; ++i) {
        cout << str[i];
    }
    cout << '\n';
}