CodeForces 1883G1 Dances (Easy version)

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-16 2.CodeForces 1935A Entertainment in MAC2024-06-16 3.CodeForces 1992C Gorilla and Permutation2024-07-16 4.CodeForces 1992D Test of Love2024-07-16 5.CodeForces 1992B Angry Monk2024-07-16 6.CodeForces 1992A Only Pluses2024-07-16 7.CodeForces 1983B Corner Twist2024-07-16 8.CodeForces 1983A Array Divisibility2024-07-16 9.CodeForces 1983C Have Your Cake and Eat It Too2024-07-16 10.CodeForces 1883B Chemistry2024-07-26 11.CodeForces 1883A Morning2024-07-26 12.CodeForces 1883C Raspberries2024-07-28 13.CodeForces 1883D In Love2024-07-28 14.CodeForces 1883E Look Back2024-07-28 15.CodeForces 1883F You Are So Beautiful2024-07-28

16.CodeForces 1883G1 Dances (Easy version)2024-07-28

17.CodeForces 908B New Year and Buggy Bot2024-08-01 18.CodeForces 908C New Year and Curling2024-08-01 19.CodeForces 1132B Discounts2024-08-01 20.CodeForces 1619D New Year's Problem2024-08-01 21.Codeforces Round 964 (Div. 4)2024-08-07

题目链接：CodeForces 1883G1【Dances (Easy version)】

思路

为了使得数组a，b中的每个对应元素满足a[i] < b[i]，所以将数组a，b按从小到大依次排列，优先删除数组a中较大的元素和数组b中较小的元素，由于删去的元素个数具有单调性，所以使用二分优化，计算最少要删去几个元素。

代码

 #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
 
const int N = 1e6 + 10;
const double eps = 1e-8;
 
ll n, m;
ll a[N], b[N];
 
bool check(int x) {
  for (int i = 1; i + x <= n; i++) {
    if (a[i] >= b[i + x]) {
      return false;
    }
  }
  return true;
}
 
void solve() {
  cin >> n >> m;
 
  a[1] = 1;
  for (int i = 2; i <= n; i++) {
    cin >> a[i];
  }
 
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
 
  sort(a + 1, a + 1 + n);
  sort(b + 1, b + 1 + n);
 
  ll l = 0, r = n, mid = 0, res = 0;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      r = mid - 1;
      res = mid;
    } else {
      l = mid + 1;
    }
  }
  cout << res << endl;
}
 
int main() {
  int t;
  cin >> t;
  while (t--)
    solve();
  return 0;
}

posted @ 2024-07-28 17:24 薛定谔的AC 阅读(10) 评论(0) 编辑收藏举报

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CodeForces 1883G1 Dances (Easy version)

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	#include <bits/stdc++.h>
	using namespace std;
	typedef long long ll;
	typedef pair<int, int> pii;

	const int N = 1e6 + 10;
	const double eps = 1e-8;

	ll n, m;
	ll a[N], b[N];

	bool check(int x) {
	for (int i = 1; i + x <= n; i++) {
	if (a[i] >= b[i + x]) {
	return false;
	}
	}
	return true;
	}

	void solve() {
	cin >> n >> m;

	a[1] = 1;
	for (int i = 2; i <= n; i++) {
	cin >> a[i];
	}

	for (int i = 1; i <= n; i++) {
	cin >> b[i];
	}

	sort(a + 1, a + 1 + n);
	sort(b + 1, b + 1 + n);

	ll l = 0, r = n, mid = 0, res = 0;
	while (l <= r) {
	mid = (l + r) >> 1;
	if (check(mid)) {
	r = mid - 1;
	res = mid;
	} else {
	l = mid + 1;
	}
	}
	cout << res << endl;
	}

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