将数字分解成3部分即可找到最大和
给出一个数字。我们的任务是将数字三倍地除以n/2,n/3和n/4,然后将数字分为三部分来求出最大和。
例如,可以将50分为{25,16,12},现在将集合{25,16,12}中的每一个再次分成三部分,依此类推。完成除法多达3次后,我们将计算总和以找到最大的乘积。
可以以递归的方式解决该程序,但是在递归的方法中,我们需要多次查找相同的结果,因此,如果使用动态编程方法并将先前计算的数据存储在表中,则它将减少时间。
输入输出
Input: Let the given number is 12. Output: The answer is 13. At first divide the 12 as (12/2 + 12/3 + 12/4) = 6 + 4 + 3 = 13. now divide 6 into three parts (6/2 + 6/3 + 6/4) = 3 + 2 + 1 = 6. If we divide the 4 and 3, we can get maximum 4 from them. From all values the maximum is 13.
算法
maxBreakSum(n)
输入: 给定的数字。
输出:分解后的最大和。
Begin define sums array of size n+1 sums[0] := 0, sums[1] := 1 for i in range 2 to n, do sum[i] := maximum of i and (sums[i/2] + sums[i/3] + sums[i/d]) done return sums[n] End
示例
#include<iostream> #define MAX 1000000 using namespace std; int max(int a, int b) { return (a>b)?a:b; } int maxBreakSum(int n) { int sumArr[n+1]; sumArr[0] = 0, sumArr[1] = 1; //for number 0 and 1, the maximum sum is 0 and 1 respectively for (int i=2; i<=n; i++) //for number 2 to n find the sum list sumArr[i] = max(sumArr[i/2] + sumArr[i/3] + sumArr[i/4], i); //divide number by 2, 3, 4 return sumArr[n]; } int main() { int n; cout << "Enter a number: "; cin >> n; cout << "Maximum sum after breaking: " << maxBreakSum(n); }
输出结果
Enter a number: 12 Maximum sum after breaking: 13