Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2042 Accepted Submission(s): 962
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.
Author
KUT（DPRK）
Source
2015 Multi-University Training Contest 10
值得借鉴
//题意：有M元钱，N种物品，每种物品的单价为W[i]，如果都买第i件商品K件，则需要花费W[i]*K元钱，同时会获得A[i]*K+B[i]个糖果，问可以获得的最大糖果数量。 //思路：先来一遍01背包，价值看成是A[i]+B[i]，再来一遍完全背包，价值看成是A[i]即可。 #include<stdio.h> #include<string.h> #include<string.h> #include<stack> #include<queue> #include<math.h> #include<algorithm> using namespace std; const int maxn = 1005; int W[maxn]; int A[maxn]; int B[maxn]; int DP[maxn*2]; int M,N; int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&M,&N); //初始化 memset(W,0,sizeof(W)); memset(A,0,sizeof(A)); memset(B,0,sizeof(B)); memset(DP,0,sizeof(DP)); for(int i = 1; i <= N; i++) { scanf("%d%d%d",&W[i],&A[i],&B[i]); } 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]);//01背包决定取不取 } for(int j = W[i]; j <= M; j++) { DP[j] = max(DP[j],DP[j-W[i]]+A[i]); //完全背包决定取最大价值 } } printf("%d\n",DP[M]); } return 0; }
