Senior‘s Array

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 852    Accepted Submission(s): 311

Problem Description

One day, Xuejiejie gets an array A. Among all non-empty intervals of A, she wants to find the most beautiful one. She defines the beauty as the sum of the interval. The beauty of the interval---[L,R] is calculated by this formula : beauty(L,R) = A[L]+A[L+1]+……+A[R]. The most beautiful interval is the one with maximum beauty.

But as is known to all, Xuejiejie is used to pursuing perfection. She wants to get a more beautiful interval. So she asks Mini-Sun for help. Mini-Sun is a magician, but he is busy reviewing calculus. So he tells Xuejiejie that he can just help her change one value of the element of A to P . Xuejiejie plans to come to see him in tomorrow morning.

Unluckily, Xuejiejie oversleeps. Now up to you to help her make the decision which one should be changed(You must change one element).

Input

In the first line there is an integer T, indicates the number of test cases.

In each case, the first line contains two integers n and P. n means the number of elements of the array. P means the value Mini-Sun can change to.

The next line contains the original array.

1≤n≤1000, −109≤A[i],P≤109。

Output

For each test case, output one integer which means the most beautiful interval‘s beauty after your change.

Sample Input

2
3 5
1 -1 2
3 -2
1 -1 2

Sample Output

8

2

Source

BestCoder Round #47 ($)

/**
    题意：给出一个数组，如果用p代替数组中的一个数求最大的区间和
    做法：dp(比赛的时候想得太多)
**/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#define maxn 1100
#define INF 0x7fffffff
using namespace std;
long long  mmap[maxn];
long long  dp[2][maxn];
long long n;
long long DP()
{
    dp[0][0] = max((long long)0,mmap[0]);
    for(int i=1;i<n;i++)
    {
        dp[0][i] = max((long long)0,dp[0][i-1] + mmap[i]);
    }
    long long tmp = mmap[0];
    for(int i=1;i<n;i++)
    {
        dp[1][i] = dp[0][i-1] + mmap[i];
        tmp = max(tmp,dp[1][i]);
    }
    return tmp;
}
int main()
{
//#ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long p;
        scanf("%lld %lld",&n,&p);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&mmap[i]);
        }
        memset(dp,0,sizeof(dp));
        long long res = -INF;
        for(int i=0;i<n;i++)
        {
            long long tt = mmap[i];
            mmap[i] = p;
            res = max(res,DP());
            mmap[i] = tt;
        }
        printf("%lld\n",res);
    }
    return 0;
}