我們把僅僅包括因子2、3和5的數稱作醜數(Ugly Number)。比如6、8都是醜數,但14不是,由於它包括因子7。

方法1 ：

暴力破解。逐個推斷

代碼：

<pre name="code" class="cpp">#include <iostream>
#include <vector>

using namespace std;

//推斷是否是醜數
bool isUgly(int index){
		while(index % 2 == 0){
			index /= 2;
		}
		while(index % 3 == 0){
			index /= 3;
		}
		while(index % 5 ==0){
			index /=5;
		}
		if(index == 1)
			return true;
		else
			return false;
}
int  print(int index){
	int count=1;
	int number = 0;
	int uglyFound =0;
	while(uglyFound < index){
		++number;
		if(isUgly(number)){
			++uglyFound;
		}
	}
	return number;
}



int main()
{	
	cout<<print(1500);

    return 0;
}

方法2 ： 採用空間換時間，僅僅是推斷醜數。

一個醜數能夠有另外一個醜數* 2 或者*3 或者*５　得到。

#include <iostream>
#include <vector>

using namespace std;

int Min(int pm2,int pm3,int pm5){
	int min = pm2 > pm3 ? pm3 : pm2;
	return min > pm5 ?
 pm5 : min;
}

void print(unsigned int index){
	if(index == 0)
		return;
	int * pUglyNumber = new int[index];
	int pNextIndex=1;
	pUglyNumber[0] = 1;
	int *pM2 = pUglyNumber;
	int *pM3 = pUglyNumber;
	int *pM5 = pUglyNumber;
	while(pNextIndex < index){
		int min=Min(*pM2 * 2,*pM3 * 3, *pM5 * 5);
		pUglyNumber[pNextIndex] = min;
		while(*pM2 * 2 <=pUglyNumber[pNextIndex])
			++pM2;
		while(*pM3 * 3 <=pUglyNumber[pNextIndex])
			++pM3;
		while(*pM5 * 5 <= pUglyNumber[pNextIndex])
			++pM5;
		pNextIndex ++;

	}
	int ugly = pUglyNumber[pNextIndex - 1];
	delete [] pUglyNumber;
	cout<< ugly;
}




int main()
{	
	print(7);

    return 0;
}

兩種方法求醜數