母函数 HDU 1085 Holding Bin-Laden Captive!

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;

	

}