学步园

这几天看了某大神写的关于母函数的文章，总算基本搞明白，就找这类题练练，这题是母函数变种

       一开始WA了好几次，搞不懂原因， 后来发现原来我忽略了，不能最小的不能组合数 可以大于所有1,2,5的总和

#include <iostream>

using namespace std;

int main()
{
	int n;
	int i, j, k;
	int addnum;

	int num[3];
	int a[8005];
	int b[8005];

	int firstboundary;
	int secondboundary;
	bool isfind = false;

    //freopen("C:\\Users\\Haojian\\Desktop\\test.txt", "r", stdin);

	cin >> num[0] >> num[1] >> num[2];
	

	while ( (num[0] || num[1] || num[2])) 
	{
		isfind = false;
		n = num[0] + num[1] * 2 + num[2] * 5;
		//各种初始化
		for (i = 0; i <= n; i++) 
		{
			a[i] = 0;
			b[i] = 0;
		}

		for (i = 0; i <= num[0]; i++)	
			a[i] = 1;
		
		for ( i = 2; i <= num[1] * 2; i += 2)
		    a[i] = 1;

		for ( i = 5; i <= num[2] * 5; i += 5)
			a[i] = 1;

		//母函数模板
		for (i = 2; i <= 3; i++) //表示第i个多项式
		{
			//设置每次的边界，和k加的次数
			if (i == 3)
			{
				addnum = 5;
				firstboundary = num[0] + num[1] * 2 ;
				secondboundary = num[0] + num[1] * 2 + num[2] * 5;
			}
			else 
			{
				addnum = 2;
				firstboundary = num[0];
				secondboundary = num[0] + num[1]*2;
			}

			//多项式计算
			for (j = 0; j <= firstboundary; j++)
			{	
				
				for (k = 0; k + j <= secondboundary; k += addnum)//第i个多项式的每个项		
					b[j + k] += a[j];				
			}

			for (k = 0; k <= secondboundary; k++) 
			{				
				a[k] = b[k];
				b[k] = 0;
			}

		}

		//查找答案输出
		for (i = 1; i <=n; i++)
		{
			if (a[i] == 0)
			{
				isfind = true;
				cout << i << endl;
				break;
			}
		}
		//如果查找不到，意味着最小的不可能组合的数就为总数+1
		if (!isfind)
			cout << n+1 << endl;

		cin >> num[0] >> num[1] >> num[2];
		
	}
	return 0;
	
}