LeetCode 1155. 掷骰子的N种方法（DP）

示例 1：
输入：d = 1, f = 6, target = 3
输出：1

示例 2：
输入：d = 2, f = 6, target = 7
输出：6

示例 3：
输入：d = 2, f = 5, target = 10
输出：1

示例 4：
输入：d = 1, f = 2, target = 3
输出：0

示例 5：
输入：d = 30, f = 30, target = 500
输出：222616187
 
提示：
1 <= d, f <= 30
1 <= target <= 1000

class Solution {	// C++
public:
    int numRollsToTarget(int d, int f, int target) {
    	vector<vector<int>> dp(d+1,vector<int>(target+1, 0));
    	dp[0][0] = 1;
    	int i,j,k;
    	for(i = 1; i <= d; ++i)
    	{
    		for(j = 0; j < target; ++j)
    		{
    			if(dp[i-1][j] != 0)//上一次的状态存在
    			{
    				for(k = 1; k <= f; ++k)
    				{
    					if(j+k <= target)//状态转移
    						dp[i][j+k] = (dp[i][j+k]+dp[i-1][j])%1000000007;
    				}
    			}
    		}
    	}
    	return dp[d][target];
    }
};

class Solution:# py3
    def numRollsToTarget(self, d: int, f: int, target: int) -> int:
        dp = [[0 for i in range(target+1)] for i in range(d+1)]
        dp[0][0] = 1
        for i in range(1,d+1):
            for j in range(0, target):
                if dp[i-1][j]==0:
                    continue
                for k in range(1, f+1):
                    if j+k <= target:
                        dp[i][j+k] = (dp[i][j+k]+dp[i-1][j])%1000000007
        return dp[d][target]