Ugly Numbers
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 150′th ugly number.
METHOD 1 (Simple)
Thanks to Nedylko Draganov for suggesting this solution.
Algorithm:
Loop for all positive integers until ugly number count is smaller than n, if an integer is ugly than increment ugly number count.
To check if a number is ugly, divide the number by greatest divisible powers of 2, 3 and 5, if the number becomes 1 then it is an ugly number otherwise not.
For example, let us see how to check for 300 is ugly or not. Greatest divisible power of 2 is 4, after dividing 300 by 4 we get 75. Greatest divisible power of 3 is 3, after dividing 75 by 3 we get 25. Greatest divisible power of 5 is 25, after dividing 25 by 25 we get 1. Since we get 1 finally, 300 is ugly number.
下面是这个简单方法，这里带了可以打印所有小于n的ugly number的程序：
int maxDivide(int num, int div)
{
while (num % div == 0)
{
num /= div;
}
return num;
}
bool isUgly(int num)
{
num = maxDivide(num, 2);
num = maxDivide(num, 3);
num = maxDivide(num, 5);
return num == 1? true:false;
}
int getNthUglyNo(int n)
{
int c = 0;
int i = 0;
while (c < n)
{
if (isUgly(++i)) c++;
}
return i;
}
#include <vector>
using std::vector;
vector<int> getAllUglyNo(int n)
{
vector<int> rs;
for (int i = 1; i <= n; i++)
{
if (isUgly(i)) rs.push_back(i);
}
return rs;
}
动态规划法：
注意： 可能会有重复的数字要跃过，程序注释出去了。否则答案错误！
细想一下，其实一句话概括就是：
由最小的Ugly number算起，然后每个都分别乘以2,3,5得到最终答案。
class UglyNumbers
{
public:
int getNthUglyNo(int n, vector<int> &rs)
{
if (n < 1) return n;
int n2 = 2, n3 = 3, n5 = 5;
int i2 = 0, i3 = 0, i5 = 0;
rs.resize(n, 1);
for (int i = 1; i < n; i++)
{
int t = min(n2, min(n3,n5));
if (t == n2)
{
rs[i] = n2;
n2 = rs[++i2]*2;
}
if (t == n3) //注意：不能使用else，避免重复
{
rs[i] = n3;
n3 = rs[++i3]*3;
}
if (t == n5) //注意：不能使用else
{
rs[i] = n5;
n5 = rs[++i5]*5;
}
}
return rs.back();
}
};
测试：
int main()
{
unsigned no = getNthUglyNo(35);
printf("ugly no. is %d \n", no);
vector<int> rs = getAllUglyNo(35);
for (auto x:rs) cout<<x<<" ";
cout<<endl;
UglyNumbers un;
printf("Ugly no. is %d \n", un.getNthUglyNo(35, rs));
for (auto x:rs) cout<<x<<" ";
cout<<endl;
system("pause");
return 0;
}
