CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

const int N = 2010;

int dp[N];
int p[N],w[N],k[N];
int n,m;
int flag[N][N];

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

    return 0;
}

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