hdu 5586 Sum 2015.11.28 bestcoder 1002

题意：给你一个函数f（x），一个数列，你有一次操作机会，能操作数列的一个区间，让区间里所有数都变成f（x），也可以不变。问操作完之后最大子段和是多少。

思路：求出每个f（x）和 x 的差，有正有负，题目就变成了这个序列的最大子段和问题，O（n）扫一遍就搞定了。

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 138    Accepted Submission(s): 92

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)

 

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#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <iomanip>
#include <time.h>
#include <set>
#include <map>
#include <stack>
using namespace std;
typedef long long LL;
const int INF=0x7fffffff;
const int MAX_N=10000;

int n;
int A[100009];
int fa[100009];

int f(int x){
    return (1890*x+143)%10007;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        int sum=0;
        int sum2=0;
        int cursum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&A[i]);
            sum+=A[i];
            fa[i]=f(A[i])-A[i];
            if(fa[i]>0){
                cursum+=fa[i];
                sum2=max(sum2,cursum);
            }
            else{
                cursum+=fa[i];
                if(cursum<0){
                    cursum=0;
                }
            }
        }
        printf("%d\n",sum+sum2);
    }
    return 0;
}