hdu-5586 Sum(dp)

题目链接：

Sum

Time Limit: 2000/1000 MS (Java/Others)  

  Memory Limit: 65536/65536 K (Java/Others)

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≤10^5)
Next line contains n integers A1,A2....An.(0≤Ai≤10^4)
It's guaranteed that ∑n≤10^6.

 

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：

 

给一个数组,选一个区间[l,r]把a[i]变成f(a[i]),也可以不选,问最后得到的这个数组的和最大是多少;

 

思路：

 

p[i]=(1890*a[i]+143)%10007;

然后就是在p[i]中找一个连续喝最大的字串,然后就好了;水题;

 

AC代码：

//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e14;
const int N=1e4+15;
const int maxn=18;

int p[10*N],a[10*N];
LL dp[10*N];
int main()
{
    int n;
    while(cin>>n)
    {
        LL sum=0,ans=0;
        Riep(n)read(a[i]),p[i]=(1890*a[i]+143)%10007-a[i],sum=sum+a[i];
        dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            if(dp[i-1]+p[i]<0)dp[i]=0;
            else dp[i]=dp[i-1]+p[i];
            ans=max(ans,dp[i]);
        }
        cout<<sum+ans<<"
";
    }

        return 0;
}