二分答案

#include<iostream>
using namespace std;
const int N=1e6+7;
int a[N],n,k;
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++) 
        cin>>a[i];
	int l=1,r=N,mid=0,ans;
	while(l<=r){
		mid=(r+l)/2;
		int tmp=0;
		for(int i=1;i<=n;i++)
			tmp+=(a[i]/mid);
		if(tmp<k) r=mid-1;
		else l=mid+1,ans=mid;
	}
	cout<<ans<<endl;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+5;
int a[N],n,k,sum=0,l=0,r,ans;
bool check(int x){
	int g=0;
	for(int i=1;i<=n;i++){
		g+=a[i]/x;
	}
	return g>=k;
}
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	sort(a+1,a+n+1);
	r=sum/k;
	if(a[1]<k){
		cout<<0;
		return 0;
	}
	if(a[1]==k){
		cout<<a[1];
		return 0;
	}
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(mid)){
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	}
	cout<<l-1;
	return 0;
}

二分

#include<iostream>
using namespace std;
#define ll long long
const int N = 1e7 + 5;
ll n, k, Max;
ll a[N];
bool check(ll x) {
	ll cnt = 0;
	for (ll i = 1; i <= n; ++i) {
		cnt += a[i] / x;
	}
	return cnt >= k;
}
int main() {
	scanf("%lld%lld", &n, &k);
	for (ll i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
		if (a[i] > Max)Max = a[i];
	}
	ll l = 0, r = Max, mid;
	while (l < r) {
		mid = (l + r) >> 1;
		if (check(mid))l = mid + 1;
		else r = mid;
	}
	printf("%lld\n", r - 1);
	return 0;
}

#include<iostream>

using namespace std;
typedef long long LL;
const int N=1e8+10;
int n,k;
LL a[N];
bool check(int x){
	int m=0;
	for(int i=0;i<n;i++){
		m+=a[i]/x;//a[i]/x表示的是当前木棍能切出的长度为x的块数，把所有木棍的对应值加起来就是当前x的最大切出来的块数 
	}
	if(m>=k){
			return true;
	}
	return false;
}
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	int l=0,r=N;
	while(l<r){
		int mid=(l+r+1)/2;
		if(check(mid)) l=mid;//如果当前的最大值满足大于等于k，说明l还有可能更大，所以往右边找 
		else r=mid-1;
	}
	cout<<l<<endl;
	return 0;
}
 
 

#include<bits/stdc++.h>

using namespace std;

long long n, k;

long long a[1000005];

bool check(long long x) {

long long cnt = 0;
for (int i = 0; i < n; i++) {
	cnt += a[i] / x;
}
return cnt >= k;

}

int main() {

cin >> n >> k;
for (int i = 0; i < n; i++) cin >> a[i];
long long l = 0, r = 100000001;
while (l  < r) {
	long long mid = (l + r+1)>>1;
	if (check(mid)) l = mid;
	else r = mid-1;
}
cout << l << endl;
return 0;

}

嗯，经典二分，找到条件就可以了

#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e6+7;

int n,m,ans,a[N];

bool check(int x){
	int k=0;
	for(int i=0;i<n;i++){
		k+=a[i]/x;
		if(k>=m)return true;
	}
	return false ;
}

int find(int l,int r){
	int mid;
	while(l<=r){
		mid=(l+r)>>1;
		if(cheak(mid))l=mid+1;
		else r=mid-1;
	}
	return l-1;
}

int main(){
	cin>>n>>m;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	sort(a,a+n);
	ans=find(0,10000007);
	cout<<ans;
	return 0;
}