1. 问题描述

如果有三种硬币，2元，5元，7元，如何用最少的数量拼成27元？

2. 思路

（以后在更新吧）

3. 代码

//
// Created by Administrator on 2021/7/20.
//

#ifndef C__TEST01_COINDP_HPP
#define C__TEST01_COINDP_HPP

#include "iostream"
#include <vector>
using namespace std;

class CoinDP {
public:
    CoinDP(vector<int> An, int M);
    vector<int> algorithmDP(vector<int> &A, int &M);
private:
    int M;
    vector<int> A;
};

CoinDP::CoinDP(vector<int> An, int M):
A(An),M(M)
{
    A.resize(An.size());
}

vector<int> CoinDP::algorithmDP(vector<int> &A, int &M){
    vector<int> f;
    f.resize(M+1);
    int i = 0;
    int j = 0;
    f[0] = 0; //Initialization

    for(i = 1; i < f.size(); ++i){
        f[i] = INT_MAX;
        // the types of coins
        for (j = 0; j < A.size(); ++j) {
            //1. money is positive
            //2. if f[i-A[j]] is INT_MAX, f[i] must be INT_MAX
            //3. i-A[j] = i - 2, i - 5, i - 7
            if(i-A[j] >= 0 && f[i-A[j]] != INT_MAX){
                f[i] = min(f[i], f[i-A[j]] + 1);
            }
        }
    }
    if(f[M] == INT_MAX){
        f[M] = -1;
    }
    return f;
}
#endif //C__TEST01_COINDP_HPP

在main函数里验证一下：

#include <iostream>
using namespace std;
#include "CoinDP.hpp"
//动态规划问题：2元，5元，7元硬币，凑成27元

int main() {
    vector<int> An = {2,5,7};
    int M = 27;
    CoinDP cdp(An, M);

    vector<int> f;
    f = cdp.algorithmDP(An, M);

    for(vector<int>::iterator it = f.begin(); it!=f.end();it++){
        cout << *it << " ";
    }
    return 0;
}

结果如图：

硬币问题