# 1155.Number-of-Dice-Rolls-With-Target-Sum

## 1155. Number of Dice Rolls With Target Sum

## 题目地址

<https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/>

## 题目描述

```
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
```

## 代码

### Approach #1

```java
class Solution {
  int MOD = 1000_000_000 + 7;
  Map<String, Integer> memo = new HashMap<>();
  public int numRollsToTarget(int d, int f, int target) {
    if (d == 0 && target == 0) {
      return 1;
    }
    if (d > target || d * f < target) {
      return 0;
    }
    String str = d + " " + target;
    if (memo.conainsKey(str)) {
      return memo.get(str);
    }
    int res = 0;
    for (int i = 1; i <= f; i++) {
      if (i <= target) {
        res = (res + numRollsToTarget(d - 1, f, target - i)) % MOD;
      } else {
        break;
      }
    }
    memo.put(str, res);
    return res;
  }

}
```

### Approach #2 Dynamic Bottom up

```java
class Solution {
    public int numRolsToTarget(int d, int f, int target) {
        int MOD = (int)Math.pow(10, 9) + 7;
        long[][] dp = new long[d + 1][target + 1];
        dp[0][0] = 1;
        for (int i = 1; i <= d; i++) {
            for (int j = 0; j <= target; j++) {
                for (int k = 1; k <= f; k++) {
                    if (j >= k) {    // dp[i][0] == 0
                        dp[i][j] = (dp[i][j] + dp[i - 1][j - k]) % MOD;
                    } else {
                        break;
                    }
                }
            }
        }

        return (int)dp[d][target];
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/dynamic-programming/backpack-problem/1155.number-of-dice-rolls-with-target-sum.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.