CF 898 div4 E. Building an Aquarium

CF 898 div4 E. Building an Aquarium，题目大意：t组数据，每组数据第一行n,m，表示有n个水位，可填充m个水位，求最高填充多高的水位

time limit per test：2 seconds
memory limit per test：256 megabytes
input：standard input
output：standard output

Problem Statement

You love fish, that’s why you have decided to build an aquarium. You have a piece of coral made of n columns, the i-th of which is ai units tall. Afterwards, you will build a tank around the coral as follows:

• Pick an integer h ≥1— the height of the tank. Build walls of height h on either side of the tank.
• Then, fill the tank up with water so that the height of each column is h, unless the coral is taller than h; then no water should be added to this column.

For example, with a=[3,1,2,4,6,2,5] and a height of h=4, you will end up using a total of w=8 units of water, as shown.

You can use at most x units of water to fill up the tank, but you want to build the biggest tank possible. What is the largest value of h you can select?

Input

The first line contains a single integer t(1≤t≤10^4 ) — the number of test cases.

The first line of each test case contains two positive integers n and x (1≤n≤2*10^5; 1≤x≤10 ^9) — the number of columns of the coral and the maximum amount of water you can use.

The second line of each test case contains n space-separated integers ai (1≤ai≤10^9) — the heights of the coral.

The sum of n over all test cases doesn’t exceed 2*10^5.

Output

For each test case, output a single positive integer h (h≥1) — the maximum height the tank can have, so you need at most x units of water to fill up the tank.

We have a proof that under these constraints, such a value of h always exists.

SAMPLES

input

5
7 9
3 1 2 4 6 2 5
3 10
1 1 1
4 1
1 4 3 4
6 1984
2 6 5 9 1 8
1 1000000000
1

output

4
4
2
335
1000000001

Note

The first test case is pictured in the statement. With h=4 we need 8 units of water, but if h is increased to 5 we need 13 units of water, which is more than x=9 . So h=4 is optimal.

In the second test case, we can pick h=4 and add 3 units to each column, using a total of 9 units of water. It can be shown that this is optimal.

In the third test case, we can pick h=2 and use all of our water, so it is optimal.

解题过程

我的解题过程是一步一步来的，首先先写出一个for循环来判断是否全部元素一样，如第二个样例输入；再考虑，还有两种情况：第一就是可使用的水位不够如第一个样例，第二是可使用的水位充足可以再次循环如最后一个样例。

上代码

#include
using namespace std;
#define int long long
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        int s[n];
        for(int i=1; i<=n; i++)
            cin>>s[i];
        sort(s+1,s+1+n);
        //for(int i=1;i<=n;i++)
        //    cout<
        int k=1;
        int ans=s[1];
        int term=0;
        int next=0;
        while(1)
        {
            int first=s[1];
            k=1;
            int flag=0;
            for(int i=2; i<=n; i++)
            {
                if(s[i]==s[i-1])
                {
                    k++;
                }
                else
                {
                    term=s[i-1];
                    next=s[i];
                    flag=1;
                    //cout<
                    break;
                }
            }
            if(flag)
            {
                if(k*(next-first)<=m)
                {
                    m-=k*(next-first);
                    for(int i=1;i<=k+1;i++)
                        s[i]=next;
                    ans=next;
                }
                else
                {
                    cout<<term+m/k<<endl;
                    break;
                }
            }
            if(k==n)
                break;
        }

        if(k==n)
        {
            ans+=m/n;
            cout<<ans<<endl;
        }



    }
}

总结

主要是解题的思路与步骤，需要一步一步根据样例来写，注意要开longlong。