CRB and His Birthday
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1655 Accepted Submission(s): 782
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
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.
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
21HintCRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
题目大意：用m元去买商品，商品有n种，买商品的时候会附加赠送糖果，并且给出了各个商品需要的话费wi元，还有关于赠送糖果量的相关系数ai和bi。随着该物品的购买量x会赠送ai*x+bi个糖果。问，用m元如何购买商品能够获得的最多糖果数目。
解题思路：乍一看就是个完全背包，但是有额外增加的条件。随着购买量增加的赠送的糖果量。bi后面没有涉及到x。所以确定第一个的赠送量就是ai+bi。后面的赠送购买赠送的糖过量就和bi没有关系了，就变成了一个完全背包。刚开始的明显可以通过01背包来解决了。
源代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<deque>
#include<map>
#include<set>
#include<algorithm>
#include<string>
#include<iomanip>
#include<cstdlib>
#include<cmath>
#include<sstream>
#include<ctime>
using namespace std; int w[1005];
int dp[2005];
int a[1005];
int b[1005]; int main()
{
int t;
int m,n;//m钱款，n代表商品种类数
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
memset(w,0,sizeof(w));
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(i = 0; i < n; i++)
{
scanf("%d%d%d",&w[i],&a[i],&b[i]);
}
for(i = 0; i < n; i++)
{
//先跑一次01背包，这样就不用再考虑bi，因为买与不买都会赠送bi糖果，
//0次或者1次考虑完之后，就剩下多次的，再跑完全背包
for(j = m; j >= w[i]; j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+a[i]+b[i]);
}
for(j = w[i]; j <= m; j++)
{
dp[j]=max(dp[j],dp[j-w[i]]+a[i]);
}
}
printf("%d\n",dp[m]);
}
return 0;
}