[蓝桥杯]真题讲解：冶炼金属（暴力+二分）

一、视频讲解

视频讲解

二、暴力代码

//暴力代码
#include
#define endl '\n'
#define deb(x) cout << #x << " = " << x << '\n';
#define INF 0x3f3f3f3f
using namespace std;
const int N = 1e4 + 10;

void solve()
{
	int n; cin >> n;
	vector<int>a(n), b(n);//创建数组
	for(int i = 0; i < n; i ++)
		cin >> a[i] >> b[i];

	for(int i = 1; i <= 1e6; i ++)//枚举的转化率V
	{
		bool flag = true;//标记一下当前的V是否合法

		for(int j = 0; j < n; j ++)
		{
			if(b[j] != (a[j] / i))
			{
				flag = false;
				break;
			}
		}
		if(flag)
		{
			cout << i << " ";
			break;
		}
	}
	
	for(int i = 1e6; i >= 1; i --)
	{
		bool flag = true;
		for(int j = 0; j < n; j ++)
		{
			if(b[j] != (a[j] / i))
			{
				flag = false;
				break;
			}
		}
		if(flag)
		{
			cout << i << " ";
			break;
		}
	}

	
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	t = 1;
	//cin >> t;
	while(t--)
	solve();
}

三、正解代码

//冶炼金属：二分
#include
#define int long long
using namespace std;
const int N = 1e4 + 10;
int a[N], b[N];
int n;

bool check_min(int mid)
{
	for(int i = 0; i < n; i ++)
	{
		if(b[i] < a[i] / mid)
			return false;
	}

	return true;
}

bool check_max(int mid)
{
	for(int i = 0; i < n; i ++)
		if(b[i] > a[i] / mid)
			return false;

	return true;
}

void solve()
{
	cin >> n;
	for(int i = 0; i < n; i ++)
		cin >> a[i] >> b[i];

	//求最小值。
	int lmin = 1, rmin = 1e9;
	
	while(lmin < rmin)
	{
		int mid = lmin + rmin >> 1;
		if(check_min(mid))
			rmin = mid;
		else
			lmin = mid + 1;
	}

	//求最大值。
	int lmax = 1, rmax = 1e9;
	while(lmax < rmax)
	{
		int mid = lmax + rmax + 1 >> 1;
		if(check_max(mid))
			lmax = mid;
		else
			rmax = mid - 1;
	}

	cout << lmin << " " << lmax << endl;
}

signed main()
{
	ios::sync_with_stdio(0);
	int t = 1;
	// cin >> t;
	while(t--)
	solve();
}