CodeForces 441B Valera and Fruits

本文介绍了一个关于收集果实的问题,Valera需要在有限时间内从n棵果树中收集尽可能多的果实,每棵树上的果实只能在特定的两天内收集。文章提供了一段C++代码,用于解决如何最大化收集到的果实数量。

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链接:http://codeforces.com/problemset/problem/441/B

Valera and Fruits

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Valera loves his garden, where n fruit trees grow.

This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).

Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?

Input

The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.

Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.

Output

Print a single integer — the maximum number of fruit that Valera can collect.

Sample test(s)

Input
2 3
1 5
2 3
Output
8
Input
5 10
3 20
2 20
1 20
4 20
5 20
Output
60

Note

In the first sample, in order to obtain the optimal answer, you should act as follows.

  • On the first day collect 3 fruits from the 1-st tree.
  • On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
  • On the third day collect the remaining fruits from the 2-nd tree.

In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.


直接附上AC代码:


#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxn = 3005;
int n, get, a, b;
int tree[maxn];

int main()
{
	ios::sync_with_stdio(false);
	while (~scanf("%d%d", &n, &get))
	{
		memset(tree, 0, sizeof(tree));
		int maxa = 0;
		while (n--)
		{
			scanf("%d%d", &a, &b);
			tree[a] += b;
			maxa = max(maxa, a);
		}
		int sum = 0;
		for (int i=1; i<=maxa+1; i++)
		{
			if (tree[i] > get)
			{
				sum += get;
				tree[i] -= get;
				if (tree[i] > get)
					tree[i+1] += get;
				else
					tree[i+1] += tree[i];
			}
			else
				sum += tree[i];
		}
		printf("%d\n", sum);
	}
	return 0;
}


### 关于 Codeforces 1853B 的题解与实现 尽管当前未提供关于 Codeforces 1853B 的具体引用内容,但可以根据常见的竞赛编程问题模式以及相关算法知识来推测可能的解决方案。 #### 题目概述 通常情况下,Codeforces B 类题目涉及基础数据结构或简单算法的应用。假设该题目要求处理某种数组操作或者字符串匹配,则可以采用如下方法解决: #### 解决方案分析 如果题目涉及到数组查询或修改操作,一种常见的方式是利用前缀和技巧优化时间复杂度[^3]。例如,对于区间求和问题,可以通过预计算前缀和数组快速得到任意区间的总和。 以下是基于上述假设的一个 Python 实现示例: ```python def solve_1853B(): import sys input = sys.stdin.read data = input().split() n, q = map(int, data[0].split()) # 数组长度和询问次数 array = list(map(int, data[1].split())) # 初始数组 prefix_sum = [0] * (n + 1) for i in range(1, n + 1): prefix_sum[i] = prefix_sum[i - 1] + array[i - 1] results = [] for _ in range(q): l, r = map(int, data[2:].pop(0).split()) current_sum = prefix_sum[r] - prefix_sum[l - 1] results.append(current_sum % (10**9 + 7)) return results print(*solve_1853B(), sep='\n') ``` 此代码片段展示了如何通过构建 `prefix_sum` 来高效响应多次区间求和请求,并对结果取模 \(10^9+7\) 输出[^4]。 #### 进一步扩展思考 当面对更复杂的约束条件时,动态规划或其他高级技术可能会被引入到解答之中。然而,在没有确切了解本题细节之前,以上仅作为通用策略分享给用户参考。
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