12 简单dp学习

#include
#include
#include
bool dp[100000];
using namespace std;
int n;
int main()
{
	cin >> n;
	int x = 1;
	for (int i = 1;i <= n;i++)
	{
		if (i == 5 || i == 3)
		{
			dp[i] = 1;
		}
		else if(i>5)
		{
			if (dp[i - 5] || dp[i - 3])
			{
				dp[i] = 1;
			}
		}
	}
	if (dp[n])
	{
		for (int i = 0;i <= 2000;i++)
		{
			int t1 = i * 3;
			int t2 = n - i * 3;
			if (t2 % 5 == 0)
			{
				cout << i + (t2 / 5) << endl;
				return 0;
			}
		}
	}
	else
	{
		cout << "-1" << endl;
	}
	return 0;
}

 

#include
#include
#include
bool dp[3000010];
using namespace std;
int n;
int main()
{
	cin >> n;
	int ans = 0;
	while (n != 1)
	{
		if (n % 2 == 0)
		{
			n /= 2;
			ans++;
		}
		else
		{
			n--;
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}

#include
#include
#include
int dp[1000010];
using namespace std;
int n;
int main()
{
	cin >> n;
	for (int i = 1;i <= n;i++)
	{
		if (i < 5)
		{
			dp[i] = i;
		}
		else if (i < 11)
		{
			dp[i] = min(dp[i - 1], dp[i - 5]) + 1;
		}
		else
		{
			dp[i] = min(dp[i - 1], dp[i - 5]);
			dp[i] = min(dp[i], dp[i - 11] )+1;
		}
	}
	cout << dp[n] << endl;
	return 0;
}

#include
using namespace std;
int arr[35][35];
int n, m;
int main()
{
	cin >> n >> m;
	for (int i = 1;i <= n;i++)
	{
		arr[i][1] = 1;
	}
	for (int i = 1;i <= m;i++)
	{
		arr[1][i] = 1;
	}
	for (int i = 2;i <= n;i++)
	{
		for (int j = 2;j <= m;j++)
		{
			if (i % 2 != 0 || j % 2 != 0)
			{
				arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
			}
		}
	}
	cout << arr[n][m];
	return 0;
}

#include
#include
#include
#include
using namespace std;
int n;
double m[110], d[110];
int day[110];
int main()
{
	cin >> n;
	for (int i = 1;i <= n;i++)
	{
		cin >> day[i];
	}
	m[0] = 100;
	for (int i = 1;i <= n;i++)
	{
		m[i] = max(m[i - 1], d[i - 1] / day[i] * 100);
		d[i] = max(d[i - 1], m[i - 1] * day[i] / 100);
	}
	cout <setprecision(2)<
	return 0;
}