hdu 5586 Sum【dp最大子段和】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5586

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 677    Accepted Submission(s): 358

Problem Description

There is a number sequence A1,A2....An ,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai) .f(x)=(1890x+143)mod10007 .After that,the sum of n numbers should be as much as possible.What is the maximum sum?

 

Input

There are multiple test cases.
First line of each case contains a single integer n.(1≤n≤105)
Next line contains n integers A1,A2....An .(0≤Ai≤104)
It's guaranteed that ∑n≤106 .

 

Output

For each test case,output the answer in a line.

 

Sample Input

2

10000 9999

5

1 9999 1 9999 1

 

Sample Output

19999

22033

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define MAX 100000
#define LL long long
using namespace std;
LL fun(LL x)
{
	return ((1890*x)%10007+143%10007)%10007;
}
int main()
{
	int n,m,j,i,k;
	int s[MAX],a[MAX],b[MAX];
	while(scanf("%d",&n)!=EOF)
	{
		int sun=0;
	    for(i=1;i<=n;i++)
	    {
	    	scanf("%d",&s[i]);
	    	a[i]=fun(s[i]);//储存 f(x); 
	    	b[i]=a[i]-s[i];//储存f(x)和x的差值 
	    	sun+=s[i];
	    }	
	    int sum=0;
	    int Max=0;
	    for(i=1;i<=n;i++)//最大子段和代码 
	    {
	    	if(sum<=0)
	    	sum=b[i];
	    	else
	    	sum+=b[i];
	    	Max=max(Max,sum);//求出最大子段 
	    }
	    printf("%d
",sun+Max);
	}
	return 0; 
}