Coin Change : 找到重现给定金额的多种方法

Question

Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter. For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5.

问题的Link是：https://www.geeksforgeeks.org/coin-change-dp-7/

我确实遇到了所有的解决方案。

我已经使用 Matrix 方法为此提出了解决方案：

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <climits>
#include <map>
#include <set>
#include <queue>
#include <unordered_map>
#define ll long long
//Author: Nilargha Roy (neel13)
using namespace std;
int coinchangeways(int arr[],int n,int sum)
{

        int dp[n+1][sum+1];
        memset(dp,-1,sizeof(dp));


        for(int j=1;j<n;j++)
        {
            dp[0][j]=0;
        }
        for(int i=0;i<n;i++)
        {
            dp[i][0]=1;
        }



    for(int i=1;i<n+1;i++)
    {
        for(int j=1;j<sum+1;j++)
        {

            if(arr[i-1] <= j)
                dp[i][j]=dp[i][j-arr[i-1]] + dp[i-1][j];
            else
                dp[i][j]=dp[i-1][j];

        }
    }

    return dp[n][sum];

}

int main() 
{

    ios_base::sync_with_stdio(false);
    cin.tie(0);

#ifdef __APPLE__
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif
    int n;
    cin>>n;
    int arr[n];
    for(int i=0;i<n;i++)
        cin>>arr[i];
    int sum; cin>>sum;
    cout<<coinchangeways(arr,n,sum);
}

但是，当我将数组 = {1,2,3} 和 SUM = 5 作为输入时，输出应该是：5 而我得到的是 1。任何人都可以指出其中的逻辑错误我的代码？

Answer 1

当你运行你的函数在所有 for 循环之后。您将获得如下所示的 DP。

 1   0   0  -1  -1  -1

 1   1   1   0  -1  -2

 1   1   2   1   1  -1

 -1  1   2   0   2   1

dp[3][5] = 1 这就是你得到 1 的原因。

Answer 2

使用一维数组的版本：

int coinchangeways(int arr[], int n, int sum)
{
    int dp[sum+1];
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;

    for (int i = 0; i < n; i++)
    {
        for (int j = arr[i]; j <= sum; j++)
        {
                dp[j] += dp[j - arr[i]];
        }
    }

    return dp[sum];
}

Coin Change : 找到重现给定金额的多种方法

Coin Change : Find number of ways to reproduce a given sum

c++

algorithm

dynamic-programming

coin-change