hdu 5586 Sum（dp+技巧）

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)

思路：令Ai=f（Ai）-Ai，然后求一遍最大连续子序列和就能知道最多能增加的值。

       一开始统计所有数的和ans，再加上最多增加的值就是答案了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 100006
#define inf 1e12
int n;
int a[N];
int A[N];
int main()
{
   while(scanf("%d",&n)==1){
      int ans=0;
      for(int i=0;i<n;i++){
         scanf("%d",&a[i]);
         ans+=a[i];
      }
      for(int i=0;i<n;i++){
         A[i]=(a[i]*1890+143)%10007-a[i];
         //printf("===%d
",A[i]);
      }
      int ThisSum=0,MaxSum=0;
      for(int i=0;i<n;i++){
         ThisSum+=A[i];
         if(ThisSum>MaxSum){
            MaxSum=ThisSum;
         }else if(ThisSum<0){
            ThisSum=0;
         }
      }
      ans+=MaxSum;
      printf("%d
",ans);

   }
    return 0;
}