Written test questions, loose change, greedy & dynamic programming

Topic background:

Given a series of face value change, such as 1,2,5,10,20, and then give a need to pay the money, how to use these changes to minimize the use of change of Zhang Shu?

 

1. Greedy

Applicable regulations: The multiple of the face value of the change meets the relationship of 2 times or more

 

Code:

#include <stdio.h>
#include <algorithm>
const int maxn=101;
int coins[maxn];
#define inc(i,l,r) for(int i=l;i<r;i++)
#define dec(i,r,l) for(int i=r-1;i>=l;i--)

int main(){
	int count;	//面值种数
	scanf("%d",&count);
	inc(i,0,count)	scanf("%d",&coins[i]);
	int X;		//支付金额 
	scanf("%d",&X); 
	
	int ans = 0;
	
	dec(i,count,0){
		int use = X /coins[i];
		ans += use;
		X = X - coins[i] * use;
		printf("需要面额为%d的%d张, ", coins[i], use);
		printf("剩余需要支付RMB %d.\n", X);
	}
	printf("最少需要%d张RMB\n", ans);
	return 0;
}

Test Results:

 

2. Dynamic programming

If our face value does not meet the condition of 2 times, such as 1, 2, 5, 7, 10;

When the payment amount = 14, the result of greed is 3 (10 + 2 + 2), and the correct result is 2 (7 + 7);

At this time, dynamic programming will be used. . .

 

Ideas:

1. State i can be obtained from the five states i-1, i-2, i-5, i-7, i-10, namely

dp[i]=min(dp[i]-coins[j])+1     

Where amount is the total payment and count is the number of face value

2. The boundary dp [0] = 0;

 

Code:
 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

#define inc(i,l,r) for(int i=l;i<r;i++)

int coinchange(vector<int> &coins,int amount){
	vector<int> dp;
	
	inc(i,0,amount+1)	dp.push_back(-1);
	
	dp[0]=0;
	inc(i,1,amount+1)
		inc(j,0,coins.size()){
			if(i-coins[j]>=0&&dp[i-coins[j]]!=-1){
				if(dp[i]==-1||dp[i-coins[j]]+1<dp[i])
					dp[i]=dp[i-coins[j]]+1;
			}
		} 
		
	return dp[amount];
} 

int main(){
	int amount,coin,count;//支付金额，面额，钱的种类 
	vector<int> coins;

	scanf("%d",&count);	
	inc(i,0,count)	scanf("%d",&coin),coins.push_back(coin);
	scanf("%d",&amount);
	
	printf("%d",coinchange(coins,amount));
	return 0;
}

Test Results: