题意：有n个商店，有m金钱，一个商店买x件商品需要x*w[i]的金钱，得到a[i] * x + b[i]件商品(x > 0)，问最多能买到多少件商品
01背包+完全背包：首先x == 1时，得到a[i] + b[i]，若再买得到的是a[i]，那么x == 1的情况用01背包思想，x > 1时就是在01的基础上的完全背包。背包dp没刷过专题，这么简单的题也做不出来:(
/************************************************
* Author :Running_Time
* Created Time :2015-8-20 12:35:33
* File Name :E.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e3 + 10;
const int MAXM = 2e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[MAXN], b[MAXN], w[MAXN];
int dp[MAXM];
int main(void) { //HDOJ 5410 CRB and His Birthday
int T; scanf ("%d", &T);
while (T--) {
int m, n; 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;
}