CodeForces 1983C Have Your Cake and Eat It Too

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 1983C【Have Your Cake and Eat It Too】

思路

先向上取整计算出tot/3，然后依次枚举abc三个数组取区间的前后顺序，对于每个顺序依次从前往后枚举，直到取得的区间数字之和大于等于tot/3，就对下一个数组进行枚举，直到所有数组都满足取出的区间数字之和大于等于 tot/3。

代码

 #include <bits/stdc++.h>
#include <vector>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
int a[5][N];
 
void solve() {
  ll n, sum = 0;
  cin >> n;
 
  for (int i = 1; i <= 3; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
 
  for (int i = 1; i <= n; i++) {
    sum += a[1][i];
  }
 
  ll need = (sum + 2) / 3;
  vector<int> p{1, 2, 3};
 
  do {
    vector<int> l(4), r(4);
    ll s = 0, i = 1;
    bool ok = true;
    for (auto t : p) {
      int j = i;
      while (j <= n && s < need) {
        s += a[t][j++];
      }
      ok &= (s >= need);
      l[t] = i;
      r[t] = j - 1;
      i = j;
      s = 0;
    }
    if (!ok) {
      continue;
    }
    for (int i = 1; i <= 3; i++) {
      cout << l[i] << ' ' << r[i] << ' ';
    }
    cout << endl;
    return;
  } while (next_permutation(p.begin(), p.end()));
 
  cout << "-1" << endl;
}
int main() {
    int t;
    cin >> t;
 
    while (t--) {
        solve();
    }
    return 0;
}

posted @ 2024-07-16 14:57 薛定谔的AC 阅读(16) 评论(0) 编辑收藏举报

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CodeForces 1983C Have Your Cake and Eat It Too

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	#include <bits/stdc++.h>
	#include <vector>
	using namespace std;
	#define ll long long
	const int N = 2e5 + 10;
	int a[5][N];

	void solve() {
	ll n, sum = 0;
	cin >> n;

	for (int i = 1; i <= 3; i++) {
	for (int j = 1; j <= n; j++) {
	cin >> a[i][j];
	}
	}

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

	ll need = (sum + 2) / 3;
	vector<int> p{1, 2, 3};

	do {
	vector<int> l(4), r(4);
	ll s = 0, i = 1;
	bool ok = true;
	for (auto t : p) {
	int j = i;
	while (j <= n && s < need) {
	s += a[t][j++];
	}
	ok &= (s >= need);
	l[t] = i;
	r[t] = j - 1;
	i = j;
	s = 0;
	}
	if (!ok) {
	continue;
	}
	for (int i = 1; i <= 3; i++) {
	cout << l[i] << ' ' << r[i] << ' ';
	}
	cout << endl;
	return;
	} while (next_permutation(p.begin(), p.end()));

	cout << "-1" << endl;
	}
	int main() {
	int t;
	cin >> t;

	while (t--) {
	solve();
	}
	return 0;
	}