文章标题 CRB and His Birthday

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

01背包+完全背包

#include<string.h>#include<iostream>#include<algorithm>using namespace std;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int dis[2009];        memset(dis,0,sizeof(dis));        int m,n,i,j;        scanf("%d%d",&m,&n);        int a[1009],b[1009],c[1009];        for(i=1;i<=n;i++)            scanf("%d%d%d",&a[i],&b[i],&c[i]);        for(i=1;i<=n;i++)        {            for(j=m;j>=a[i];j--)//01背包                dis[j]=max(dis[j],dis[j-a[i]]+b[i]+c[i]);            for(j=a[i];j<=m;j++)//完全背包                dis[j]=max(dis[j],dis[j-a[i]]+b[i]);        }        printf("%d\n",dis[m]);    }    return 0;}