Problem Description
Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with  M Won(currency unit).
At the shop, there are  N kinds of presents.
It costs  Wi Won to buy one present of  i-th kind. (So it costs  k ×  Wi Won to buy  k of them.)
But as the counter of the shop is her friend, the counter will give  Ai × x + Bi candies if she buys  x( x>0) presents of  i-th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤  T ≤ 20
1 ≤  M ≤ 2000
1 ≤  N ≤ 1000
0 ≤  Ai, Bi ≤ 2000
1 ≤  Wi ≤ 2000
 

Input
There are multiple test cases. The first line of input contains an integer  T, indicating the number of test cases. For each test case:
The first line contains two integers  M and  N.
Then  N lines follow,  i-th line contains three space separated integers  Wi,  Ai and  Bi.
 

Output
For each test case, output the maximum candies she can gain.
 

Sample Input
1 100 2 10 2 1 20 1 1
 

Sample Output
21
Hint
CRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
 

第一次搜多校赛的题解，好在于看题解明白了啊啊啊，好激动，而且背着敲1A！！多久没有了这种感觉了T^T

说正经的-->_--> 这题啥意思 ？完全背包中若选择某种物品，另加b[i]的价值，只加一次

开始百思不得其姐啊啊啊，怎么记录某种物品买没买啊？？再加数组太麻烦,把自己绕糊涂了orz

后来题解说先进行一次01背包再进行一次多重背包 茅塞顿开啊啊啊啊

01那次加b[i] 多重背包加只a[i] 01就相当于第一次取某种物品 多重背包 理论知识详见我上一篇博客

刚刚想到会不会两个for重复矛盾了了呢？不不不 装第一个的时候 下面完全背包的总价值一定比01背包的小 Hhhhhh

贴上我光辉的代码 怪不得过的人那么多 看了题解谁都会-->_-->

#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int dp[2005],cost[2005],a[2005],b[2005];
int main()
{
    //freopen("cin.txt","r",stdin);
    int t,m,n;
    while(cin>>t)
    {
        while(t--)
        {
            cin>>m>>n;
            memset(dp,0,sizeof(dp));
            for(int i=0;i<n;i++) cin>>cost[i]>>a[i]>>b[i];
            for(int i=0;i<n;i++)
            {
                for(int j=m;j>=cost[i];j--)
                    if(dp[j]<dp[j-cost[i]]+a[i]+b[i]) dp[j]=dp[j-cost[i]]+a[i]+b[i];
                for(int j=cost[i];j<=m;j++)
                    if(dp[j]<dp[j-cost[i]]+a[i]) dp[j]=dp[j-cost[i]]+a[i];
            }
            cout<<dp[m]<<endl;
        }
    }
    return 0;
}