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≤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)

题目链接：点击传送

思路：基础dp；

　　　dp[i][0]表示dp前边没有改成f的总和

　　　dp[i][1]表示可以继续改成f的总和；

　　　dp[i][2]表示不可以继续改成f的总和；

dp[i][0]=dp[i-1][0]+a[i];
dp[i][1]=max(dp[i-1][1]+b[i],dp[i-1][0]+b[i]);
dp[i][2]=max(dp[i-1][2]+a[i],dp[i-1][1]+a[i]);

　　　b=f(a);

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<iostream>

#include<cstdio>

#include<cmath>

#include<string>

#include<queue>

#include<algorithm>

#include<stack>

#include<cstring>

#include<vector>

#include<list>

#include<set>

#include<map>

using namespace std;

#define ll long long

#define pi (4*atan(1.0))

#define eps 1e-14

#define bug(x)  cout<<"bug"<<x<<endl;

const int N=1e5+,M=1e6+,inf=;

const ll INF=1e18+,mod=;

ll a[N],b[N],dp[N][];

int main()

{

    int n;

    while(~scanf("%d",&n))

    {

        for(int i=;i<=n;i++)

            scanf("%d",&a[i]),b[i]=(*a[i]+)%;

        dp[][]=dp[][]=dp[][]=;

        for(int i=;i<=n;i++)

        {

            dp[i][]=dp[i-][]+a[i];

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

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

        }

        printf("%lld\n",max(dp[n][],max(dp[n][],dp[n][])));

    }

    return ;

}

巴特西

hdu 5586 Sum 基础dp

Sum

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