Senior‘s Array

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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 ($)

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题意：给你一个数值 改变数值中某个元素的值（a[i] = p）求改变后数组的最大子串和 必须改

思路：就是数据比较坑 如果不用__int64会被heck掉

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
__int64 a[1005];
__int64 sum[1005];
__int64 ssum[1005];

int main()
{
    __int64  t;
    scanf("%I64d",&t);
    while(t--)
    {
        __int64 n,p;
        scanf("%I64d%I64d",&n,&p);
        a[0] = 0;
        sum[0] = 0;

        __int64 Mx=-999999999;

        for(__int64 i = 0; i < n; i++)
            scanf("%I64d",&a[i]);

        for(__int64 i = 0; i < n; i++)
        {
            __int64 tp = a[i];
            a[i] = p;
             for (__int64 j = 0;j < n;j++)
            {
                if (j)
                sum[j] = a[j] > sum[j-1]+a[j] ? a[j] : sum[j-1]+a[j];
                else
                    sum[j]=a[j];
                if (sum[j]>Mx)
                    Mx=sum[j];
//                cout<<sum[j]<<endl;
            }
            a[i] = tp;
        }
        printf("%I64d\n",Mx);
    }
}

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