1. 題目
示例 1:
輸入:d = 1, f = 6, target = 3
輸出:1
示例 2:
輸入:d = 2, f = 6, target = 7
輸出:6
示例 3:
輸入:d = 2, f = 5, target = 10
輸出:1
示例 4:
輸入:d = 1, f = 2, target = 3
輸出:0
示例 5:
輸入:d = 30, f = 30, target = 500
輸出:222616187
提示:
1 <= d, f <= 30
1 <= target <= 1000
複制
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/number-of-dice-rolls-with-target-sum
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2. 解題
- 從上一次的所有狀态推導目前次的狀态
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)//上一次的狀态存在
{
for(k = 1; k <= f; ++k)
{
if(j+k <= target)//狀态轉移
dp[i][j+k] = (dp[i][j+k]+dp[i-1][j])%1000000007;
}
}
}
}
return dp[d][target];
}
};
複制
36 ms 8.7 MB
python3 注意二維數組的寫法
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]
複制
524 ms 13.9 MB