L-Flowers【2019瀋陽區域賽】【二分答案】

傳送門

應邀碼：ae17b8c95ce5aac5 

Recently Jack becomes much more romantic. He would like to prepare several bunches of flowers.

Each bunch of flowers must have exactly M flowers. As Jack does not want to be boring, he hopes that flowers in the same bunch are all different species. Now there are N species of flowers in the flower shop, and the number of the i-th species of flower is a​i​​. Now Jack would like to know how many bunches of flowers he can prepare at most.

(Flowers are used to propose.)

Input

The first line contains an integer T (1≤T≤10) --- the number of test cases.

In the first line of each test case, there are two integers N, M (1≤N,M≤300000) --- the number of flowers' species and the number of flowers in a bunch.

In the second line of each test case, there are N integers --- the i-th integer indicates a​i​​ (1≤a​i​​≤10​9​​), the number of i-th species' flowers.

Output

For each test case, output one integer in one line --- the answer of the corresponding test case.

Sample Input

1
5 3
1 1 1 2 1

Sample Output

2

 

題意：有n種花，每種花有a[i]個，求最多能組成幾個花束（每個花束m朵花，且同一個花束中沒有相同種類的花）

解題思路：這道題可用二分法解決。假設能組成x個花束，則當a[i]大於x時花i最多用x個，當a[i]小於x時，花i可以全部用上 。把這些能用到的花全部都加起來得到sum，問題就在於，這些能用到的花能組成幾束花？我們可以思考，這sum多花很有可能不會剛好組成x束花，會有多餘的，而剛好組成x束花的那部分一定是符合條件的。（如果你認爲會有相同的花在同一束裏，那你是把多餘的花和那x束花混在一塊了）這sum朵花裏，如果滿足sum>=x*m,那x就是有效的。理想情況下x的最大值是num(花的總數)/m,那麼我們只要以sum>=x*m爲判斷條件，從0到num/m進行二分尋找答案就行了。

 

#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
const int maxn=3e5+7;

int n,m;
ll a[maxn];

bool check(ll x)
{
	ll sum=0;
	for(ll i=1;i<=n;i++) sum+=min(a[i],x);
	if(sum>=x*m) return 1;
	else return 0;
}

int main()
{
//	freopen("input.txt","r",stdin);
	
	int t;
	cin>>t;
	while(t--)
	{
		ll num=0;
		cin>>n>>m;
		for(ll i=1;i<=n;i++) 
		{
			cin>>a[i];
			num+=a[i];
		}
		ll L=0,R=num/m,mid,ans;//從0開始，是因爲x的值是有可能爲0的
		while(L<=R)
		{
			mid=(L+R)/2;
			if(check(mid))
			{
				L=mid+1;
				ans=mid;
			}
			else R=mid-1;
		}
		cout<<ans<<endl;
	}
	
	return 0;
}