Question

我正在构建一个<label id="lblpendingcount" class="labels">Levelwise Approval Pending Counts</label>，它返回输入更改的硬币数。但是，我想修改代码，以便该函数还返回一个数组，其值为为示例返回的硬币：

recursive bottom-up dynamic programming function

我尝试打印Change:9c; #coins:3; 1c: 0, 2c:2, 5c:1，但只从硬币元组中提取了一枚硬币。例如：

在values[j]，可能的元组是iteration 7，但我只得到每个元素的最后一个元素，即1,2，而我想要最后一个元组。

以下是此问题的C代码：

(5,1,1),(5,2)

Answer 1

您没有随时可以获得的信息。

如果您为每次迭代存储最后一枚硬币，则可以打印硬币：

/**
  * @desc returns the number of coins needed for a particular amount of change
  *       using bottom-up recursive dynamic programming approach.
  * @param int amt  - amount of change to convert to coins.
  *        int values[]  - array of coins available (predefined).
  *        int n  - length of values[].
  *        int a  - current iteration of change (predefined at 1).
  *        int coins[]  - array holding the amount of coins from 1 till amt.
  * @return return coins[amt] - returns the amount of coins.
*/
int ComputeChange(int amt,int values[],int n,int a,int coins [], int lastcoin [])
{
    printf("\n\na: %d",a);
    printf("\n");
    coins[a]=INT_MAX;
    int j;
    int tmp;
    //loops all values of coins
    for (j=0; j<n; j++)
    {

        // if current coin value is smaller or equal than a (the current change value)
        // and if the number of coins at the current amount-the currently looped value of a coin
        // is less than the number of coins at the current amount.
        if ((values[j] <=a)&& (1+coins[a-values[j]]<coins[a]))
        {
            lastcoin[a] = values[j];
            coins[a]=1+coins[a-values[j]];
        }
    }
    if (a==amt)
    {
        j = amt;
        while (j>0) {
            printf("%d, ", lastcoin[j]);  // Print the coins that make up the amount 
            j -= lastcoin[j];
        }
        printf("\n");
        return coins[amt];
    }
    else
    {
        return ComputeChange(amt,values,n,a+1,coins);
    }
}

从动态编程功能中获取值

1 个答案: