原题链接在这里:https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/
题目:
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
题解:
Let dp[k] denotes the way to make up target k.
For each dice, there could be value j within [1, f]. dp[k - j] is accumlated to new value.
New value need all the original values of dp, thus here we need a temp.
Time Complexity: O(d * f * target).
Space: O(target).
AC Java:
1 class Solution { 2 public int numRollsToTarget(int d, int f, int target) { 3 int mod = 1000000007; 4 int [] dp = new int[target + 1]; 5 dp[0] = 1; 6 7 for(int i = 0; i < d; i++){ 8 int [] temp = new int[target + 1]; 9 for(int j = 1; j <= f; j++){ 10 for(int k = j; k <= target; k++){ 11 temp[k] = (temp[k] + dp[k - j]) % mod; 12 } 13 } 14 15 dp = temp; 16 } 17 18 return dp[target]; 19 } 20 }
类似Coin Change 2.