UVA - 136 Ugly Numbers

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, 15, ...

shows the first 11 ugly numbers. By convention, 1 is included.

Write a program to find and print the 1500'th ugly number.

Input and Output

There is no input to this program. Output should consist of a single line as shown below, with <number> replaced by the number computed.

Sample output

The 1500'th ugly number is <number>.

第一种方法：遍历每个数，判断是否为ugly，直到第1500个ugly为止（简单粗暴，没有效率可言，runtime 19s）

#include <cstdio>

using namespace std;

int isugly(int n)
{
while(n>1)
{
if(n%2==0)n/=2;
else if(n%3==0)n/=3;
else if(n%5==0)n/=5;
else return 0;
}
return 1;
}

int main()
{
int cnt = 1;
int number = 2;
while(cnt < 1500)
{
if(isugly(number))cnt++;
number++;
}
printf("%d\n",number-1);
return 0;
}

第二种方法：利用STL优先队列从小到大生成各个ugly number。最小的丑数是1，而对于任意丑数x，2x，3x和5x也都是丑数。注意丑数判重。（有点难度，效率高的没话说，runtime 5ms）

#include <iostream>
#include <vector>
#include <queue>
#include <set>

using namespace std;

typedef long long LL;
const int coeff[3] = {2, 3, 5};

int main()
{
priority_queue<LL, vector<LL>, greater<LL> > pq;
set<LL> s;
pq.push(1);
s.insert(1);
for(int i=1; ;i++)
{
LL x = pq.top(); pq.pop();
if(i==1500)
{
cout << "The 1500'th ugly number is " << x << ".\n";
break;
}
for(int j=0; j<3; j++)
{
LL x2 = x*coeff[j];
if(!s.count(x2)){s.insert(x2); pq.push(x2);}
}
}
}