2019 ICPC沈阳 Flowers

题意： 给定$n$种花，第$i$种有$a_i$朵，一束花由$m$朵互不相同种类的花组成，问最多可以组成多少束花。
数据范围：$1\le n,m\le 3\times 10^5,1\le a_i\le 10^9$

题解： 二分答案，设当前答案为$x$，判断答案是否合法，合法则真实答案至少为$x$，否则真实答案至多为$x-1$。判断条件是每种花取$c_i=min(x,a_i)$，最后判断$sum\ge \sum c_i$是否成立即可。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
using namespace std;

typedef long long ll;

template<typename T>
inline T Read(){
    
    
    T s = 0, f = 1; char ch = getchar();
    while(!isdigit(ch)) {
    
    if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {
    
    s = (s << 3) + (s << 1) + ch - '0'; ch = getchar();}
    return s * f;
}

#define read() Read<int>()
#define readl() Read<long long>()

const int N = 3e5 + 10;
ll a[N];
int n, m;

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

void solve() {
    
    
	n = read(), m = read();
	ll sum = 0;
	for(int i = 1; i <= n; i++) a[i] = readl(), sum += a[i];
	
	ll l = 0, r = sum / m;
	while(l < r) {
    
    
		ll mid = l + r + 1 >> 1;
		if(check(mid)) l = mid;
		else r = mid - 1;
	}
	printf("%lld\n", l);
}

int main()
{
    
    
	int T = 1;
	T = read();
	for(int i = 1; i <= T; ++i) {
    
    
		solve();
	}

	return 0;
}