hdu1085（普通母函数）

最新推荐文章于 2019-06-29 14:22:25 发布

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本文介绍了一个基于硬币问题的算法挑战，由本·拉登提出，涉及中国硬币（1元、2元、5元）的数量限制，目标是找出无法支付的最小金额。通过使用母函数模板和动态规划技巧，文章提供了一种有效解决该问题的方法。

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Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26528 Accepted Submission(s): 11681

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

解析：母函数模板

#include<bits/stdc++.h>
using namespace std;

#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

const int maxn=8e3+5;
struct node{
	int val,cnt;
}num[3];
int c1[maxn],c2[maxn];

int main()
{
	num[0].val=1,num[1].val=2,num[2].val=5;
	while(~scanf("%d%d%d",&num[0].cnt,&num[1].cnt,&num[2].cnt))
	{
		if(num[0].cnt==0 && num[1].cnt==0 && num[2].cnt==0)break;
		
		int mav=num[0].cnt*num[0].val+num[1].cnt*num[1].val+num[2].cnt*num[2].val;
		mem(c1,0);
		mem(c2,0);
		for(int i=0; i<=num[0].cnt; i++)
		{
			c1[i]=1;
		}
		for(int i=1; i<3; i++)
		{
			for(int j=0; j<=mav; j++)
			{
				for(int k=0; k+j<=mav && k<=num[i].cnt*num[i].val; k+=num[i].val)
				{
					c2[j+k]+=c1[j];
				}
			}
			for(int j=0; j<=mav; j++)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		int i;
		for(i=1; i<=mav; i++)
		{
			if(!c1[i])break;
		}
		
		printf("%d\n",i);
	}
	return 0;
}