让我们考虑一个游戏,在该游戏中,玩家每步动作可以得到3、5或10的分数。还给出目标分数。我们的任务是找到用这三个点来达到目标分数的可能方法。
通过动态编程方法,我们将创建一个从0到n的所有得分的列表,并且对于3、5、10的每个值,我们只需更新表即可。
Input: The maximum score to reach using 3, 5 and 10. Let the input is 50. Output: Number of ways to reach using (3, 5, 10)50: 14
countWays(n)
只有3分,分别是3、5和10
输入: n是要达到的最高分数。
输出-达到分数n的可能方法的数量。
Begin create table of size n+1 set all table entries to 0 table[0] := 1 for i := 3 to n, do table[i] := table[i] + table[i-3] done for i := 5 to n, do table[i] := table[i] + table[i-5] done for i := 10 to n, do table[i] := table[i] + table[i-10] done return table[n] End
#include <iostream> using namespace std; //返回达到分数n的方法数量 int countWay(int n) { int table[n+1], i; //table to store count for each value of i for(int i = 0; i<=n; i++) { table[i] = 0; // Initialize all table values as 0 } table[0] = 1; //set for 1 for input as 0 for (i=3; i<=n; i++) //try to solve using 3 table[i] += table[i-3]; for (i=5; i<=n; i++) //try to solve using 5 table[i] += table[i-5]; for (i=10; i<=n; i++) //try to solve using 10 table[i] += table[i-10]; return table[n]; } int main() { int n; cout << "Enter max score: "; cin >> n; cout << "Number of ways to reach using (3, 5, 10)" << n <<": " << countWay(n); }
输出结果
Enter max score: 50 Number of ways to reach using (3, 5, 10)50: 14