CodeForces 1992E Novice's Mistake

题目链接：CodeForces 1992E 【Novice's Mistake】

思路

       直接对a,b枚举肯定会超时，因为a,b数数字过大，但是通过结果a * n - b可以发现结果最多为6位数，所以对结果的位数进行枚举，然后枚举a，来计算出b并判断是否符合题意，同时需要去掉b不符合题目的范围的情况。

代码

#include <functional>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stdio.h>
#include <string>
#include <cstring>
#include <vector>
using namespace std;
#define ll long long
const int N = 50090 + 10;

struct ab {
  ll a, b;
} res[N];

void solve(){
  ll n, a, b, ans = 1;
  cin >> n;
  string s1 = to_string(n);
  for (int i = 2; i <= 10000; i++) {
    string s2 = "";
    while (s2.length() <= 10)
      s2 += s1;

    for (int j = 1; j <= 6; j++) {
      ll b = s1.size() * i - j;
      if (b < 1 || b > min(10000ll, n * i))
        continue;
      
      if (to_string(i * n - b) == s2.substr(0, j)) {
        res[ans++] = {i, b};
	  }
	}
  }
  cout << ans - 1 << endl;
  for (int i = 1; i < ans; i++) {
    cout << res[i].a << " " << res[i].b << endl;
  }
}

int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}