Flowers（二分答案）

链接：https://ac.nowcoder.com/acm/contest/7830/J
来源：牛客网

题目描述

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{N}N species of flowers in the flower shop, and the number of the i-th species of flower is aia_iai. Now Jack would like to know how many bunches of flowers he can prepare at most.

(Flowers are used to propose.)\textit{(Flowers are used to propose.)}(Flowers are used to propose.)

输入描述:

The first line contains an integer T{T}T (1≤T≤101 \le T \le 101≤T≤10) --- the number of test cases.
In the first line of each test case, there are two integers N{N}N, M{M}M (1≤N,M≤300 0001 \le N, M \le 300\,0001≤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{N}N integers --- the i{i}i-th integer indicates aia_iai (1≤ai≤1091 \le a_i \le 10^91≤ai≤109), the number of i{i}i-th species' flowers.

输出描述:

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

示例1

输入

复制1 5 3 1 1 1 2 1

1
5 3
1 1 1 2 1

输出

复制2

题意：

有n种花，然后要求每束花种至少有m种不同种类的花，问最多能有多少束花

二分答案：

通过先推答案来找是否符合要求

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<time.h>
#define ll long long
using namespace std;
ll a[300100];
ll b[300100];
int n,m;ll check(ll mid){//判断是否符合要求ll p = 1ll*mid*m;ll sum = 0;for(ll i = 1; i <= n; i++){sum += min(1ll*a[i],mid);//如果小于mid（即答案）则其必然能被使用完全//大于mid的最多只能使用mid个}return sum >= p;}
int main(){ll sum = 0;ll t;scanf("%lld",&t);while(t--){//memset(a,0,sizeof(a));cin >> n >> m;sum=0;for(int i = 1; i <= n; i++){scanf("%lld",&a[i]);sum += a[i];}sort(a+1,a+1+n);//特例：如果n > m ，则不能完成要求//相同则输出最小的数if(m > n){cout << 0 << endl;}else if(m == n){cout << a[1] << endl;}else{//从0到1e14中寻找答案ll l = 0;ll r = 1e14;ll ans = 0;while(l <= r){ll mid = (l+r)/2;ll num = check(mid);if(num == 1){l = mid+1;ans= mid;}else{r = mid-1;}}printf("%lld\n",ans);}}return 0;
}