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

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#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\n",Loli+sum);         }         else         {             printf("%I64d\n",sum);         }     }     return 0; }