Educational Codeforces Round 130 B. Promo题目链接题目详情 The stor

题目链接

题目详情

The store sells n items, the price of the i-th item is $p_i$ . The store's management is going to hold a promotion: if a customer purchases at least x items, y cheapest of them are free.

The management has not yet decided on the exact values of x and y. Therefore, they ask you to process q queries: for the given values of x and y, determine the maximum total value of items received for free, if a customer makes one purchase.

Note that all queries are independent; they don't affect the store's stock.

Input The first line contains two integers $n$ and $q (1≤n,q≤2⋅10^5)$ — the number of items in the store and the number of queries, respectively.

The second line contains n integers $p_1,p_2,…,p_n (1≤pi≤10^6)$ , where pipi — the price of the ii-th item.

The following q lines contain two integers $x_i$ and $y_i$ each $(1≤y_i≤x_i≤n)$ — the values of the parameters x and y in the i-th query.

Output For each query, print a single integer — the maximum total value of items received for free for one purchase.

input

output

8
5
6

题目大意即解题思路

大意：
这家商店出售n种商品，第种商品的价格是pi。这家商店的管理层将举办一场促销活动：如果顾客购买了至少x件商品，y最便宜的商品是免费的。管理层尚未决定x和y的确切价值。因此，他们要求你处理q查询：对于给定的x和y值，如果客户进行一次购买，则确定免费收到的物品的最大总价值。请注意，所有查询都是独立的；它们不会影响商店的库存。每次查询给定x和y对于每个查询，打印一个整数-一次购买免费获得的项目的最大总价值。
解题思路：
首先我们会想到先将整个数组排序，然后从后往前就可以枚举我们要买的x个商品的价格，进而我们也可以找到这个区间里的最便宜的y项商品的价格，然后求和。但是我们看到数据量q和n都是 $2·10^5$ 时，就可以判断出会超时的（~~第一次做没看到~~），果不其然运行后会超时！！！然后这里求得又是某段区间的和，所以我们首先想到前缀和，这里我们只需找到区间的两个端点，，然后我们就可以降低时间复杂度，AC题目。

超时代码：

 while (q--)
    {
        int x, y;
        cin >> x >> y;
        int sum = 0;
        for (int i = n - x; i < n - x + y ; i++)
        {
            sum += a[i];
        }
        cout << sum << endl;
    }

AC代码

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
long long a[200010], s[200010] ;
int main()
{
    int n, q;
    cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];
    sort(a+1, a + n+1);

    for (int i = 1; i <= n; i++) 
    {
        s[i] = s[i - 1] + a[i];
    }
    while (q--)
    {
        int x, y;
        cin >> x >> y;
        long long sum = s[n - x + y ] - s[n - x];
        
        cout << sum << endl;
    }
    return 0;
}