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

题意：给定背包容量m，物品种类n，每种物品的重量w[i]，还有卖的数量x对应的价值a[i]*x+b[i]，求最大的价值。

解法：01背包选好第一个物品，即确定价值量a[i]+b[i]，然后完全背包走一遍，每次价值加a[i]。

#include <bits/stdc++.h>
using namespace std;
int T, m, n, w[1010], a[1010], b[1010];
int dp[2010];

int main()
{
    scanf("%d", &T);
    while(T--){
        scanf("%d %d", &m,&n);
        for(int i=1; i<=n; i++) scanf("%d %d %d", &w[i],&a[i],&b[i]);
        memset(dp, 0, sizeof(dp));
        for(int i=1; i<=n; i++){
            for(int j=m; j>=w[i]; j--){
                dp[j] = max(dp[j], dp[j-w[i]]+a[i]+b[i]);
            }
            for(int j=w[i]; j<=m; j++){
                dp[j] = max(dp[j], dp[j-w[i]]+a[i]);
            }
        }
        printf("%d\n", dp[m]);
    }
    return 0;
}