Codeforces 1250J The Parade

题目链接

The Berland Army is preparing for a large military parade. It is already decided that the soldiers participating in it will be divided into k rows, and all rows will contain the same number of soldiers.

Of course, not every arrangement of soldiers into k rows is suitable. Heights of all soldiers in the same row should not differ by more than 1. The height of each soldier is an integer between 1 and n.

For each possible height, you know the number of soldiers having this height. To conduct a parade, you have to choose the soldiers participating in it, and then arrange all of the chosen soldiers into k rows so that both of the following conditions are met:

each row has the same number of soldiers,
no row contains a pair of soldiers such that their heights differ by 2 or more.
Calculate the maximum number of soldiers who can participate in the parade.

Input

The first line contains one integer t (1≤t≤10000) — the number of test cases. Then the test cases follow.

Each test case begins with a line containing two integers n and k (1≤n≤30000, 1≤k≤1e12) — the number of different heights of soldiers and the number of rows of soldiers in the parade, respectively.

The second (and final) line of each test case contains n integers c1, c2, …, cn (0≤ci≤1e12), where ci is the number of soldiers having height i in the Berland Army.

It is guaranteed that the sum of n over all test cases does not exceed 30000.

Output

For each test case, print one integer — the maximum number of soldiers that can participate in the parade.

Example

input

5
3 4
7 1 13
1 1
100
1 3
100
2 1
1000000000000 1000000000000
4 1
10 2 11 1

output

16
100
99
2000000000000
13

二分，题意是问一共有多少士兵可以参加游行，又要求必须要 k k k 行，那么我们可以二分一行的最大人数，因为要求每一行的人的高度差不超过 1 1 1，我们就直接贪心取，按高度从低到高取，取满 k k k 行为止，AC代码如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=3e4+5;
int t,n;
ll a[N],k,b[N];
bool check(ll x){
    ll u=k;
    for(int i=1;i<=n;i++){
        b[i]=a[i];
    }
    for(int i=1;i<=n;i++){//贪心取人
        ll cnt=b[i]/x;
        b[i]=b[i]%x;
        if(u>=cnt) u-=cnt;
        else return 1;
        if(b[i+1]+b[i]>=x){
            b[i+1]-=(x-b[i]);
            b[i]=0;
            u--;
        }
        if(u<=0) return 1;
    }
    return (u<=0);
}

int main()
{
    cin>>t;
    while(t--){
        memset(b,0,sizeof(b));
        cin>>n>>k;
        ll sum=0;
        for(int i=1;i<=n;i++) cin>>a[i],sum+=a[i];
        ll l=1,r=sum/k+1;
        while(l<=r){
            ll mid=l+r>>1;
            if(check(mid)) l=mid+1;
            else r=mid-1;
        }
        cout<<r*k<<endl;
    }
    return 0;
}