Senior's Array（5280）

最新推荐文章于 2025-07-16 07:00:00 发布

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最新推荐文章于 2025-07-16 07:00:00 发布

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本文介绍了一个算法问题，即通过仅改变数组中的一个元素来最大化选定子数组的总和。通过遍历并尝试替换每个元素，找到能产生最大总和的最佳方案。

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*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

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

Sample Output

8
2

注：暴力破解，依次替换一个数，但是要注意max的初始值要设的足够小。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <stack>
#include <set>
#include <map>
#include <string>
using namespace std;
#define pi acos(-1,0)
#define INF 2147483647
int max(int a,int b)
{
	return a>=b?a:b;
}
long long int s[1005];
int main()
{
	long long int sum,max;
	int i,j,n,p,t;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			memset(s,0,sizeof(s));
			
			scanf("%d %d",&n,&p);
			for(i=0;i<n;i++)
			scanf("%I64d",&s[i]);
			max=-INF;
			for(i=0;i<n;i++)
			{
				sum=0;
				for(j=0;j<n;j++)
				{
					if(j==i)
					sum+=p;
					else
					sum+=s[j];
					
					if(sum>max)
					max=sum;
					if(sum<0)
					sum=0;
				}
			}
			printf("%I64d\n",max);
		}
	}
	
	return 0;
}