hdu 5586 Sum 最大子段和

Sum

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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

HINT

题意

给你n个数，你可以使得一个区间的数由a[i]变成b[i]，b[i] = (1890*a[i]+143)mod10007，问你答案最大为多少

题解：

将b[i]减去a[i]，然后跑一个最大子段和就好了

可以dp，也可以尺取法

代码：

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
#define maxn 100005
long long a[maxn];
long long b[maxn];
long long sumb[maxn];
long long get(long long x)
{
    return (1890*x+143)%10007;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        long long sum = 0;
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            sum += a[i];
        }
        for(int i=1;i<=n;i++)
        {
            b[i]=get(a[i])-a[i];
        }
        long long Tmp = 0 , tmp = 0;
        for(int i = 1 ; i <= n ; ++ i)
        {
            if(tmp + b[i] < 0)
            {
                Tmp = max(Tmp , b[i]);
                Tmp = max(Tmp , tmp);
                tmp = 0;
            }
            else
            {
                tmp += b[i];
                Tmp = max(Tmp,tmp);
            }
        }
        printf("%I64d
",sum+Tmp);
    }
}