TOJ 1753.Hamming Problem

最新推荐文章于 2017-08-01 10:02:00 发布

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最新推荐文章于 2017-08-01 10:02:00 发布

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分类专栏： Toj水题数论文章标签： toj 水题数论

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本文介绍了一种通过编程解决Hamming序列问题的方法。Hamming序列包含所有仅由给定三个质数作为质因数的自然数，并按升序排列。文章提供了一个C语言程序示例，用于计算指定位置的Hamming数。

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题目链接：http://acm.tju.edu.cn/toj/showp1753.html

1753. Hamming Problem

Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 535 Accepted Runs: 251 Multiple test files

For each three prime numbers p1, p2 and p3, let's define Hamming sequence Hi(p1, p2, p3), i=1, ... as containing in increasing order all the natural numbers whose only prime divisors are p1, p2 or p3.

For example, H(2, 3, 5) = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, ...

So H5(2, 3, 5)=6.

Input

In the single line of input there are space-separated integers p1 p2 p3 i.

Output

The output must contain the single integer - Hi(p1, p2, p3). All numbers in input and output are less than 10¹⁸.

Sample Input

7 13 19 100

Sample Output

26590291

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关于丑数的一道题，丑数相关知识见：http://blog.youkuaiyun.com/acm_zl/article/details/10613073

知道怎么求丑数，这道题就很简单了：

#include <stdio.h>
long long Min(long long a,long long b,long long c){
	if(a<b)
		return a<c?a:c;
	else
		return b<c?b:c;
}
int main()
{ 
	int num2,num3,num5,n;
	while(~scanf("%d%d%d%d",&num2,&num3,&num5,&n)){
		const int Max=n;
		long long Prime[Max+1];
		int index2=0,index3=0,index5=0;
		Prime[0]=1;
		for(int i=0;i<n;i++){
			long long minnum=Min(Prime[index2]*num2,Prime[index3]*num3,Prime[index5]*num5);
			if(minnum==Prime[index2]*num2)
				index2++;
			if(minnum==Prime[index3]*num3)
				index3++;
			if(minnum==Prime[index5]*num5)
				index5++;	
			Prime[i+1]=minnum;
		}
		printf("%lld\n",Prime[n]);
	}
}