動態規劃：對角線行走

沿着對角線，重點是發現狀態轉移方程：dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

#include <iostream>
#include <vector>
using namespace std;
int main()
{
	int n=1, m=2;
	while (	cin >> n >> m)
	{
		//int dp[m][n];
		n = n + 1;
		m = m + 1;
		vector<vector<int> > dp(m , vector<int>(n , 0));
		//d[i][j]=d[i-1][j]+d[i][j-1];
		for (int i = 0; i <n; i++)//行填充,n
		{
			dp[0][i] = 1;
		}
		for (int i = 0; i < m; i++)//列填充,m
		{
			dp[i][0] = 1;
		}
		
		for (int i = 1; i <m; i++)
		{
			for (int j = 1; j < n; j++)
			{
				dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
			}
		}
		cout << dp[m-1][n-1] << endl;
	}
	return 0;
}