思路

字符串构造

s.length() <= 1e5, 如果”111…337” or “133777…”“, max_s.length() = 1e5 - 3 << 1e9
如果”133777…333…7” max_s.length() = (1e5 - 2) * (1e5 - 3) / 2 > 1e9
“1333…7” 可构造1337: i * (i - 1) / 2 <==> C(i, 2) 个
用C(i, 2) <= n模拟出i的值如果i < n 在前面补(n - i)个”7”

拓展

1. 括号序列

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int t;

int main()
{
    cin >> t;
    while(t -- )
    {
        int n;
        cin >> n;
        int m = 1;
        while(m * (m - 1) / 2 <= n) m ++ ;
        m -- ;
        cout << "133";
        for(int i = 0; i < n - m * (m - 1) / 2; i ++ ) cout << "7";
        for(int i = 0; i < m - 2; i ++ ) cout << "3";
        cout << "7" << endl;
    }
    return 0;
}