HDU5586：Sum（最大连续字段和）

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1490    Accepted Submission(s): 758

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)

思路：将每个数的f(x)-x存到新的数组中，求其中的最大连续字段和即可。

# include <bits/stdc++.h>
# define f(x) (1890*x+143)%10007

using namespace std;
int main()
{
    int n, a, b, imin, imax, sum, sum2;
    while(~scanf("%d",&n))
    {
        imin = sum = imax = sum2 = 0;
        while(n--)
        {
            scanf("%d",&a);
            b = f(a) - a;
            sum += a;
            sum2 += b;
            imax = max(imax, sum2-imin);
            imin = min(imin, sum2);
        }
        printf("%d\n",sum+imax);
    }
    return 0;
}

下面是idealism_xxm的dp代码：

#include <cstdio>
#include <algorithm>

using namespace std;

int n,num,i,dp[2][3];//dp[i&1][0]表示一直未进入置换区间，dp[i&1][1]表示当前处于置换区间，dp[i&1][2]表示当前已出置换区间（前面已经进入过了）

inline int f() {
    return (1890*num+143)%10007;
}

int main() {
    while(1==scanf("%d",&n)) {
        dp[0][0]=dp[0][1]=dp[0][2]=0;
        for(i=1;i<=n;++i) {
            scanf("%d",&num);
            dp[i&1][0]=dp[(i-1)&1][0]+num;
            dp[i&1][1]=max(dp[(i-1)&1][0],dp[(i-1)&1][1])+f();
            dp[i&1][2]=max(dp[(i-1)&1][1],dp[(i-1)&1][2])+num;
        }
        printf("%d\n",max(dp[n&1][0],max(dp[n&1][1],dp[n&1][2])));
    }
    return 0;
}