Girl Love Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Love in college is a happy thing but always have so many pity boys or girls can not find it.

Now a chance is coming for lots of single boys. The Most beautiful and lovely and intelligent girl in HDU,named Kiki want to choose K single boys to travel Jolmo Lungma. You may ask one girls and K boys is not a interesting thing to K boys. But you may not
know Kiki have a lot of friends which all are beautiful girl!!!!. Now you must be sure how wonderful things it is if you be choose by Kiki.

Problem is coming, n single boys want to go to travel with Kiki. But Kiki only choose K from them. Kiki every day will choose one single boy, so after K days the choosing will be end. Each boys have a Love value (Li) to Kiki, and also have a other value (Bi),
if one boy can not be choose by Kiki his Love value will decrease Bi every day.

Kiki must choose K boys, so she want the total Love value maximum.

Input

The input contains multiple test cases.

First line give the integer n,K (1<=K<=n<=1000)

Second line give n integer Li (Li <= 100000).

Last line give n integer Bi.(Bi<=1000)

Output

Output only one integer about the maximum total Love value Kiki can get by choose K boys.

Sample Input

3 3
10 20 30
4 5 6
4 3
20 30 40 50
2 7 6 5

Sample Output

47
104

Author

yifenfei

Source

HDU女生专场公开赛——谁说女子不如男

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题目大意：有n个男孩，每个男孩对那个女孩都有一个爱慕值li，一个递减值bi，当女孩不选他时他的爱慕之减去没选他的天数乘递减值，每天只能选一个男孩，问选m个获得最大的m值是多少

ac代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
struct s
{
	int li,bi;
}a[1010];
int dp[1010];
int cmp(const void *a,const void *b)
{
	return (*(struct s *)b).bi-(*(struct s *)a).bi;
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i].li);
		}
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i].bi);
		}
		qsort(a,n,sizeof(a[0]),cmp);
		memset(dp,0,sizeof(dp));
		for(i=0;i<n;i++)
		{
			for(j=m;j>=1;j--)
			{
				dp[j]=max(dp[j],dp[j-1]+a[i].li-(j-1)*a[i].bi);
			}
		}
		printf("%d\n",dp[m]);
	}
}

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