HDOJ/HDU 1085 母函数 Holding Bin-Laden Captive!

最新推荐文章于 2019-06-29 14:22:25 发布

最新推荐文章于 2019-06-29 14:22:25 发布 · 343 阅读

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#java

本文揭示了拉登在杭州隐匿期间，为避免外出而沉迷于解决数学问题的故事。具体问题涉及使用不同面额的中国硬币组合支付最小无法支付的价值。通过模拟多项式乘法，作者轻松解决了此问题，并分享了解题过程及代码实现。

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6600Accepted Submission(s): 2894

Problem Description

We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
“Oh, God! How terrible! ”

Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

Input

Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.

Output

Output the minimum positive value that one cannot pay with given coins, one line for one case.

Sample Input

1 1 3
0 0 0

Sample Output

4

这个题我们仍然模拟多项式乘法。最后计算就是了。

我的代码：

#include<stdio.h>
#include<stdlib.h>

int c1[10000];
int c2[10000];

int main()
{
	int i,j,num1,num2,num3,sum;
	while(scanf("%d%d%d",&num1,&num2,&num3)!=EOF)
	{
		if(num1==0&&num2==0&&num3==0)
			break;
		sum=num1+2*num2+5*num3;
		for(i=0;i<=sum;i++)
		{
			c1[i]=0;
			c2[i]=0;
		}
		for(i=0;i<=num1;i++)
			c1[i]=1;
		for(i=0;i<=num1;i++)
			for(j=0;j<=2*num2;j=j+2)
				c2[i+j]=c2[i+j]+c1[i];
		for(i=0;i<=num1+2*num2;i++)
		{
			c1[i]=c2[i];
			c2[i]=0;
		}
		for(i=0;i<=num1+2*num2;i++)
			for(j=0;j<=5*num3;j=j+5)
				c2[i+j]=c2[i+j]+c1[i];
		for(i=0;i<=sum;i++)
		{
			c1[i]=c2[i];
			c2[i]=0;
		}
		bool flag=true;
		for(i=1;i<=sum;i++)
			if(c1[i]==0)
			{
				flag=false;
				printf("%d\n",i);
				break;
			}
		if(flag)
			printf("%d\n",sum+1);
	}
	return 0;
}