time limit per test
2 seconds
memory limit per test
256 megabytes
The store sells n items, the price of the i-th item is pi. 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⋅105) --- the number of items in the store and the number of queries, respectively.
The second line contains n integers p1,p2,...,pn (1≤pi≤106), where pi --- the price of the i-th item.
The following q lines contain two integers xi and yi each (1≤yi≤xi≤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.
Example
Input
Copy
5 3
5 3 1 5 2
3 2
1 1
5 3Output
Copy
8
5
6Note
In the first query, a customer can buy three items worth 5,3,5, the two cheapest of them are 3+5=8.
In the second query, a customer can buy two items worth 5 and 5, the cheapest of them is 5.
In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is 1+2+3=6.
解题说明：此题是一道模拟题，采用贪心算法。首先根据物品价格进行排序，然后先计算出包含前面所有物品的总价。最后根据输入的查询条件直接输出结果即可。
cpp
#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define ll long long
int main()
{
ll n, q, x, y, i;
cin >> n >> q;
ll p[n + 1]; p[0] = 0;
for (i = 0; i < n; i++)
{
cin >> p[i + 1];
}
sort(p, p + n + 1);
for (i = 1; i <= n; i++)
{
p[i] += p[i - 1];
}
while (q--)
{
cin >> x >> y;
cout << (p[n + y - x] - p[n - x]) << endl;
}
return 0;
}