CodeForces 1992E Novice's Mistake

1.CodeForces 1992E Novice's Mistake2024-07-16

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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;
}

posted @ 2024-07-16 13:00 薛定谔的AC 阅读(32) 评论(0) 编辑收藏举报

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CodeForces 1992E Novice's Mistake

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	#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;
	}