CodeForces 1883E Look Back

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 1883E【Look Back】

思路

若直接对每个元素进行操作累乘至大于相邻的前一个元素时，可能最后会数据溢出，而且乘的2个数可能会很多，会时间超限。所以可以对每两个相邻的元素进行判断，判断他们之间差了2的多少次方。cnt记录的是当前元素和上个元素之间差的2的cnt次方。如：数组a为1, 4, 1, 4, 9，i = 2时，count = -2，cnt = max(0, 0 + -2)，i = 3时，count = 2，cnt = max(0, 0 + 2)，此时cnt表示a[3]和a[2]之间差了2的cnt次方，后续在cnt的基础上操作代表上个元素需要乘2的cnt次方，此时直接将当前元素和上个元素比较原元素的大小关系加上之前求出的cnt就可以得到当前元素需要乘2的cnt次方，i = 4时，count = -2，cnt = max(0, 2 + -2)，i = 5时，count = -2，cnt = max(0, 0 + -2)。

代码

 #include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 10;
ll a[N];
void solve() {
  ll n, res = 0;
  cin >> n;
 
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
 
  ll cnt = 0, ans = 0;
  for (int i = 2; i <= n; i++) {
    ll start = a[i - 1], target = a[i], count = 0;
    // 计算当前元素与上一个元素之间的差值
    while (start < target)
      count--, start *= 2;
    while (start > target)
      count++, target *= 2;
	// 综合上个元素的乘2次数，计算当前元素乘2的次数
    cnt = max(0ll, count + cnt);
    res += cnt;
  }
 
  cout << res << endl;
}
 
int main() {
  int t;
  cin >> t;
 
  while (t--) {
    solve();
  }
  return 0;
}

posted @ 2024-07-28 15:44 薛定谔的AC 阅读(6) 评论(0) 编辑收藏举报

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CodeForces 1883E Look Back

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	#include <bits/stdc++.h>
	using namespace std;
	#define ll long long
	const int N = 1e5 + 10;
	ll a[N];
	void solve() {
	ll n, res = 0;
	cin >> n;

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

	ll cnt = 0, ans = 0;
	for (int i = 2; i <= n; i++) {
	ll start = a[i - 1], target = a[i], count = 0;
	// 计算当前元素与上一个元素之间的差值
	while (start < target)
	count--, start *= 2;
	while (start > target)
	count++, target *= 2;
	// 综合上个元素的乘2次数，计算当前元素乘2的次数
	cnt = max(0ll, count + cnt);
	res += cnt;
	}

	cout << res << endl;
	}

	int main() {
	int t;
	cin >> t;

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