​

题意

codeforces1328B_字符串组数据，每次给你两个数codeforces1328B_ios_02，问你由codeforces1328B_字符串_03个codeforces1328B_ios_04和codeforces1328B_字符串_05个codeforces1328B_ios_06组成的字符串的第codeforces1328B_#define_07小的字典序的字符串是什么。


思路

我们通过观察可以知道规律，1+2+3+…+n，符合前codeforces1328B_字符串_08项和的部分规律，第codeforces1328B_#define_07个串由两个codeforces1328B_ios_06的位置唯一确定，由后往前找第一个codeforces1328B_ios_06的位置，这个位置就是前codeforces1328B_字符串_08项和大于codeforces1328B_#define_07的位置，第二个位置就是前codeforces1328B_ios_14项和与codeforces1328B_#define_07的差，详细见代码。

参考代码

#include <bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
const double pi = acos(-1.0);
const double eps = 1e-9;
typedef long long LL;
const int N = 1e6 + 10;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double f = 2.32349;
void solve() {
  IOS;
  int t;
  cin >> t;
  while (t--) {
    LL n, k;
    cin >> n >> k;
    int x = 0, y = 0;
    while (1) {
      y++;
      x = y * (y + 1) / 2;
      if (x >= k) {
        break;
      }
    }
    x = y * (y - 1) / 2;
    x = k - x;
    cout << x << " " << y << endl;
    y = n - y;
    x = n + 1 - x;
    cout << x << " " << y << endl;
    for (int i = 1; i <= n; i++) {
      if (i == x || i == y) {
        cout << "b";
      } else {
        cout << "a";
      }
    }
    cout << endl;
  }
}
signed main() {
  solve();
  return 0;
}