POJ1338 & POJ2545 & POJ2591 & POJ2247

时间：2015-07-16 12:03:25 阅读：134 评论：0 收藏：0 [点我收藏+]

标签：poj

POJ1338 2545 2591 2247都是一个类型的题目，所以放到一起来总结

POJ1338：Ugly Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21708		Accepted: 9708

Description

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n‘th ugly number.

Input

Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.

Output

For each line, output the n’th ugly number .:Don’t deal with the line with n=0.

Sample Input

Sample Output

1
2
10

找质因子为2、3、5的数，实际这些数就是2、3、5这些数互相乘，从大到小排好序的序列。

发现这种题也有一个固定的套路，也渐渐知道模板题是个什么概念了。

代码：

#include <iostream>
using namespace std;

int main()
{
	int a[1500]={1},i=1,j2=0,j3=0,j5=0,m,count;
	while(i<1500)
	{
		m=999999999;
		if(m>2*a[j2])m=2*a[j2];
		if(m>3*a[j3])m=3*a[j3];
		if(m>5*a[j5])m=5*a[j5];
		if(m==2*a[j2])j2++;
        if(m==3*a[j3])j3++;
		if(m==5*a[j5])j5++;
		a[i]=m;
		i++;
	}
	while(cin>>count&&count)
	{
		cout<<a[count-1]<<endl;
	}
	return 0;
}

POJ2545：Hamming Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6560		Accepted: 3010

Description

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 file there are space-separated integers p1 p2 p3 i.

Output

The output file must contain the single integer - Hi(p1, p2, p3). All numbers in input and output are less than 10^18.

Sample Input

7 13 19 100

Sample Output

26590291

和前面的题目一个意思，一开始的问题在于输入不超过10^18，心想这不是搞笑吗，又要TLE了。求出来打表？后来发现输出也要不超过10^18。于是就试试，结果成了。

代码：

#include <iostream>
#pragma warning(disable:4996)
using namespace std;
#define MAXN 10006

long long a[MAXN];

int main()
{
	a[0] = 1;
	int i=1,i2,j1=0,j2=0,j3=0,p1,p2,p3;
	long long m;
	cin>>p1>>p2>>p3>>i2;
	while(i<=10005)
	{
		m=9223372036854775807;
		if(m>p1*a[j1]) m=p1*a[j1];
		if(m>p2*a[j2]) m=p2*a[j2];
		if(m>p3*a[j3]) m=p3*a[j3];

		if(m==p1*a[j1])j1++;
		if(m==p2*a[j2])j2++;
		if(m==p3*a[j3])j3++;
		a[i]=m;
		i++;
	}

	cout<<a[i2]<<endl;
	return 0;
}

POJ2591：Set Definition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9509		Accepted: 4465

Description

Set S is defined as follows:
(1) 1 is in S;
(2) If x is in S, then 2x + 1 and 3x + 1 are also in S;
(3) No other element belongs to S.

Find the N-th element of set S, if we sort the elements in S by increasing order.

Input

Input will contain several test cases; each contains a single positive integer N (1 <= N <= 10000000), which has been described above.

Output

For each test case, output the corresponding element in S.

Sample Input

100
254

Sample Output

418
1461

代码：

#include <iostream>
#pragma warning(disable:4996)
using namespace std;

int a[10000005];

int main()
{
	a[0] = 1;
	int i=1,j2=0,j3=0;
	long long m;
	while(i<=10000000)
	{
		m=9223372036854775807;
		if(m>2*a[j2])m=2*a[j2]+1;
		if(m>3*a[j3])m=3*a[j3]+1;
		
		if(m==2*a[j2]+1)j2++;
        if(m==3*a[j3]+1)j3++;
		a[i]=m;
		i++;
	}
	while(scanf("%d",&i)==1)
	{
		cout<<a[--i]<<endl;
	}
	return 0;
}

POJ2247：Humble Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9951		Accepted: 4651

Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.

Write a program to find and print the nth element in this sequence.

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.

Sample Input

Sample Output

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

做到这里就对这种题很烦了。。。

代码：

#include <iostream>
#pragma warning(disable:4996)
using namespace std;
#define MAXN 10006

long long a[MAXN];

int main()
{
	a[1] = 1;
	int i=1,i2,j1=1,j2=1,j3=1,j4=1;
	long long m;

	while(i<=5842)
	{
		m=4000000000;
		if(m>2*a[j1]) m=2*a[j1];
		if(m>3*a[j2]) m=3*a[j2];
		if(m>5*a[j3]) m=5*a[j3];
		if(m>7*a[j4]) m=7*a[j4];

		if(m==2*a[j1])j1++;
		if(m==3*a[j2])j2++;
		if(m==5*a[j3])j3++;
		if(m==7*a[j4])j4++;
		a[++i]=m;
	}
	while(cin>>i2)
	{
		if(!i2)
			break;
		if((i2%100)>=10&&(i2%100)<=20)
		{
			cout<<"The "<<i2<<"th humble number is "<<a[i2]<<"."<<endl;
		}
		else if(i2%10==1)
		{
			cout<<"The "<<i2<<"st humble number is "<<a[i2]<<"."<<endl;
		}
		else if(i2%10==2)
		{
			cout<<"The "<<i2<<"nd humble number is "<<a[i2]<<"."<<endl;
		}
		else if(i2%10==3)
		{
			cout<<"The "<<i2<<"rd humble number is "<<a[i2]<<"."<<endl;
		}
		else 
		{
			cout<<"The "<<i2<<"th humble number is "<<a[i2]<<"."<<endl;
		}
	}

	return 0;
}

所以总结一下的话，因为按一条一条的要求逐渐去查找，前一个数又作为查找后一个数的基础，所以有多少条件就搞多少个j1,j2,j3。取最小的那个，之后选择了哪一个条件，就将对应条件的jn+1，让它到队列的下一个，接着判断，逐渐得到一整个数的序列。

POJ1338 & POJ2545 & POJ2591 & POJ2247

标签：poj

原文地址：http://blog.csdn.net/u010885899/article/details/46906807

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