K for the Price of One（EASY HARD）

This is the easy version of this problem. The only difference is the constraint on k — the number of gifts in the offer. In this version: k=2.

Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky — today the offer “k of goods for the price of one” is held in store. Remember, that in this problem k=2.

Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.

More formally, for each good, its price is determined by ai — the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:

Vasya can buy one good with the index i if he currently has enough coins (i.e p≥ai). After buying this good, the number of Vasya’s coins will decrease by ai, (i.e it becomes p:=p−ai).
Vasya can buy a good with the index i, and also choose exactly k−1 goods, the price of which does not exceed ai, if he currently has enough coins (i.e p≥ai). Thus, he buys all these k goods, and his number of coins decreases by ai (i.e it becomes p:=p−ai).
Please note that each good can be bought no more than once.

For example, if the store now has n=5 goods worth a1=2,a2=4,a3=3,a4=5,a5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.

Help Vasya to find out the maximum number of goods he can buy.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases in the test.

The next lines contain a description of t test cases.

The first line of each test case contains three integers n,p,k (2≤n≤2⋅105, 1≤p≤2⋅109, k=2) — the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.

The second line of each test case contains n integers ai (1≤ai≤104) — the prices of goods.

It is guaranteed that the sum of n for all test cases does not exceed 2⋅105. It is guaranteed that in this version of the problem k=2 for all test cases.

Output
For each test case in a separate line print one integer m — the maximum number of goods that Vasya can buy.

Example
Input
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
Output
3
4
2
0
4
3
emmmm这道题设计的这个k不知道有啥用，可能有简单的做法吧。对于不同的k我们都可以dp来做。
对于第i件物品，我们可以有两种做法来买，一个是直接买，另一种就是买从i-k+1~i件，这两种方法取最优的。我们用dp[i]代表的是买前i件物品所需要的最小值。最后遍历一遍就可以了。
代码如下：

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;const int maxx=2e5+100;
int a[maxx];
ll dp[maxx];//dp[i]代表的是前i件物品花费最小值
int n,p,k;int main()
{int t;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&p,&k);for(int i=1;i<=n;i++) scanf("%d",&a[i]);sort(a+1,a+1+n);//2 3 4 5 7->11memset(dp,inf,sizeof(dp));dp[0]=0;for(int i=1;i<=n;i++){if(i>=k) dp[i]=min(dp[i-1],dp[i-k])+(ll)a[i];else dp[i]=dp[i-1]+(ll)a[i];}int ans=0;for(int i=1;i<=n;i++){if(dp[i]<=p) ans=i;}cout<<ans<<endl;}return 0;
}

努力加油a啊，(^o)/~

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