HDU 5586 (dp 思想)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 507    Accepted Submission(s): 284

Problem Description

There is a number sequence A1,A2....An

,A1,A2....An

,A1,A2....An

,A1,A2....An

,A1,A2....An

,A1,A2....An

,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)

 

传送门 http://acm.hdu.edu.cn/showproblem.php?pid=5586

类似杭电1003

题意  一串序列 可以只能更改一个区间 或不更改f(x)=(1890x+143)mod10007    更改操作为f（x）=（1890x+143）mod10007

询问 这串序列的和的最大值

做一个变形 每个值变为 f（x）-x 然后按照1003的方法 求最大连续区间的和 

dp[i]=max(dp[i],dp[i-1]+dp[i]);

 

f(x)=(1890x+143)mod10007

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n;
int bac(int s)
{
    return (1890*s+143)%10007;
}
int b[100005],a[100005];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        int re=0;
        for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                re=re+a[i];
                b[i]=bac(a[i])-a[i];
            }
          //  int exm=0;
            int maxn=0;
          for(int i=1;i<n;i++)
          {
              if(b[i]+b[i-1]>b[i])
                b[i]=b[i]+b[i-1];
       if(b[i]>maxn)
                     maxn=b[i];

          }
        printf("%d
",re+maxn);
    }
    return 0;
}