開始用完全揹包寫，過不去，不知道思路錯在哪兒了；

擴充套件歐幾里德定理

使用擴充套件歐幾里德演算法解決不定方程的辦法

倒咖啡

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 21   Accepted Submission(s) : 9

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Problem Description

Input

Output

Sample Input

15 3 20
5 4 4

Sample Output

YES
NO

#include <iostream>
using namespace std;

int gcd(int x, int y){return (y == 0)?x:gcd(y,x%y);}
int main()
{
	int V,VA,VB;
	while(cin>>V>>VA>>VB)
	{
		if(V%gcd(VA,VB)==0)
		cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	
}
 

我的思路是：物品有4種，價值為VA的，VB的，VA-VB的，還有就是min(VA,VB) -  max(VA,VB)%min(VA,VB);

比如VA = 3;VB = 20;

物品就有：3  30  17  1

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int f[1003],V = 1001;
int CompletePack(int cost, int worth)
{
	for(int j = cost; j <= V; j++)
	{
		f[j] = max(f[j], f[j-cost]+worth);
	}
} 

int main()
{
	int V1,VA,VB,a[5];
	while(scanf("%d%d%d",&V1,&VA,&VB)!=EOF)
	{
		a[1] = VA;
		a[2] = VB;
		a[3] = abs(VA-VB);
		a[4] =min(VA,VB) -  max(VA,VB)%min(VA,VB);
		memset(f,0,sizeof(f));
		for(int i = 1; i <= 4;i++)
			CompletePack(a[i],a[i]);
		
		if(f[V1] == V1)
		cout<<"YES"<<endl;
		else
		cout<<"NO"<<endl;
	}
}