BestCoder Round #64 (div.2) 1002 Sum HDU 5586 联机算法

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于 2015-11-28 23:18:18 发布

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本文探讨了一道经典的算法题目，即如何通过选择性转换数列中的元素来获得最大的数列和。通过定义一个转换函数f(x)，文章提供了一个有效的算法实现思路，包括联机算法的代码示例。

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Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 113 Accepted Submission(s): 74

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

Source

BestCoder Round #64 (div.2)

出题人：令 ${A}_{i}=f({A}_{i})-{A}_{i}$ ,然后求一遍最大连续子序列和就能知道最多能增加的值。

联机算法 :

#include
  
   
#include
   
    
#include
    
     
using namespace std;
int main()
{
    int n,x,s,sum,maxn,y;
    while(~scanf("%d",&n))
    {
        maxn=sum=s=0;
        for(int i=0; i
     
      sum?maxn:sum;
        }
        printf("%d\n",s+maxn);
    }
    return 0;
}