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+)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.(≤n≤)
Next line contains n integers A1,A2....An.(≤Ai≤)
It's guaranteed that ∑n≤106.

 

Output

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

Sample Input

Sample Output

 

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)==){
       int ans=;
       for(int i=;i<n;i++){
          scanf("%d",&a[i]);
          ans+=a[i];
       }
       for(int i=;i<n;i++){
          A[i]=(a[i]*+)%-a[i];
          //printf("===%d\n",A[i]);
       }
       int ThisSum=,MaxSum=;
       for(int i=;i<n;i++){
          ThisSum+=A[i];
          if(ThisSum>MaxSum){
             MaxSum=ThisSum;
          }else if(ThisSum<){
             ThisSum=;
          }
       }
       ans+=MaxSum;
       printf("%d\n",ans);

    }
     return ;
 }