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1155.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

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

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];
}
}