BestCoder Round #64 1002

Sum

 

 Accepts: 322

 

 Submissions: 940

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

 

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

Problem Description

There is a number sequenceA​1​​,A​2​​....A​n​​,you can select a interval [l,r] or not,all the numbers A​i​​(l≤i≤r) will become f(A​i​​).f(x)=(1890x+143) mod 10007f(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 (0≤A​i​​≤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

我们可以把所有的数都尝试的换一下，但是需要一个变量记录增长量，我们取增长量最大的加上初始的和就行

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
int a[100010];
int main()
{
    int n;
    int i,j;
    __int64 sum,ans,Loli;
    int b;
    while(~scanf("%d",&n))
    {
        sum=0;Loli=0;ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
     //   cout<<sum<<endl;
        for(i=0;i<n;i++)
        {
            if((a[i]*1890+143)%10007>a[i])
            {
                ans+=(a[i]*1890+143)%10007-a[i];
            }
            else if((a[i]*1890+143)%10007<=a[i])
            {
                ans-=a[i]-(a[i]*1890+143)%10007;
            }
            if(ans>Loli)
            {
                Loli=ans;
            }
           if(ans<0)
           {
               ans=0;
           }
          // cout<<ans<<endl;
        }
        if(Loli>0)
        {
            printf("%I64d
",Loli+sum);
        }
        else
        {
            printf("%I64d
",sum);
        }
    }
    return 0;
}