HDOJ 5586 Sum (最大连续子串和)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 489    Accepted Submission(s): 275

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
   
   
   
   

 

题意：n个数的序列A，可以选择一段的Ai，使其中的Ai变为f(Ai)，也可以不变（PS:原来做的时候没看到那个not，汗= =），求变化后的最大和
思路：将序列中每个数和对应的其变化过的数之差存起来，然后求得最长连续子串和 最后加上原始的序列总和

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 100010
#define INF 0xfffffff 
using namespace std;
LL num[MAXN];
int a[MAXN];
LL ans;
int main()
{
	int i,n,j;
    while(scanf("%d",&n)!=EOF)
    {
    	ans=0;
    	for(i=1;i<=n;i++)
    	{
    		scanf("%d",&a[i]);
    		ans+=a[i];
    		num[i]=(a[i]*1890+143)%10007-a[i];
		}
		int s=0,M=0;
		for(i=1;i<=n;i++)
		{
			s+=num[i];
			if(s>0)
			M=max(M,s);
			else
			s=0;
		}
		printf("%lld\n",ans+M);
	}
    return 0;
}