【leetcode】1155. Number of Dice Rolls With Target Sum

题目如下：

You have d dice, and each die has f faces numbered 1, 2, ..., f.

Return the number of possible ways (out of fd total ways) modulo 10^9 + 7 to roll the dice so the sum of the face up numbers equals target.

Example 1:

Input: d = 1, f = 6, target = 3
Output: 1
Explanation: 
You throw one die with 6 faces.  There is only one way to get a sum of 3.

Example 2:

Input: d = 2, f = 6, target = 7
Output: 6
Explanation: 
You throw two dice, each with 6 faces.  There are 6 ways to get a sum of 7:
1+6, 2+5, 3+4, 4+3, 5+2, 6+1.

Example 3:

Input: d = 2, f = 5, target = 10
Output: 1
Explanation: 
You throw two dice, each with 5 faces.  There is only one way to get a sum of 10: 5+5.

Example 4:

Input: d = 1, f = 2, target = 3
Output: 0
Explanation: 
You throw one die with 2 faces.  There is no way to get a sum of 3.

Example 5:

Input: d = 30, f = 30, target = 500
Output: 222616187
Explanation: 
The answer must be returned modulo 10^9 + 7.

Constraints:

• 1 <= d, f <= 30
• 1 <= target <= 1000

解题思路：记dp[i][j] 为前i个骰子掷完后总和为j的组合的总数。那么很显然(i-1)个骰子掷完后的总和只能是  (j-d)  ~ (j-1) ,所以有dp[i][j] = sum(dp[i-1][j-d] , dp[i-1][j-d + 1] .... + dp[i-1][j-1]) 。

代码如下：

class Solution(object):
    def numRollsToTarget(self, d, f, target):
        """
        :type d: int
        :type f: int
        :type target: int
        :rtype: int
        """
        if target > d*f:
            return 0
        dp = [[0] * (d*f+1) for i in range(d)]
        for i in range(1,f+1):
            dp[0][i] = 1

        for i in range(1,len(dp)):
            for j in range(len(dp[i])):
                for k in range(1,f+1):
                    dp[i][j] += dp[i-1][j-k]
        #print dp

        return dp[-1][target] % (10**9 + 7)