把只含因子2、3和5的数称为丑数，求按从小到大的顺序的第1500个丑数。例如6、8都是丑数，但14不是，因为它包含因子7。习惯上把1当作第1个丑数。

算法1：逐个判断每个整数是不是丑数的解法，直观但不够高效

#include


int ugly(int number)

{

while (number % 2 == 0)

{

number /= 2;

}

while (number % 3 == 0)

{

number /= 3;

}

while (number % 5 == 0)

{

number /= 5;

}

return (number == 1) ? 1 : 0;

}


int fun(int num)

{

if (num <= 0)

{

return 0;

}

int number = 0;

int uglyfound = 0;

while (uglyfound < num)

{

++number;

if (ugly(number))

{

++uglyfound;

}

}

return number;

}


int main()

{

int num = 0, ret = 0;

printf("请输入一个丑数的序号:");

scanf("%d", &num);

ret = fun(num);

printf("number=%d\n", ret);

return 0;

}

算法2：不需要在非丑数的整数上做任何计算，时间效率明显提升，但是由于需要保存已经生成的丑数，增加了空间消耗，相当于用空间消耗换取了时间效率的提升。主要代码如下：

int Min(int number1, int number2, int number3);


int GetUglyNumber_Solution(int index)

{

if (index <= 0)

return 0;


int *pUglyNumbers = new int[index];

pUglyNumbers[0] = 1;

int nextUglyIndex = 1;


int *pMultiply2 = pUglyNumbers;

int *pMultiply3 = pUglyNumbers;

int *pMultiply5 = pUglyNumbers;


while (nextUglyIndex < index)

{

int min = Min(*pMultiply2 * 2, *pMultiply3 * 3, *pMultiply5 * 5);

pUglyNumbers[nextUglyIndex] = min;


while (*pMultiply2 * 2 <= pUglyNumbers[nextUglyIndex])

++pMultiply2;

while (*pMultiply3 * 3 <= pUglyNumbers[nextUglyIndex])

++pMultiply3;

while (*pMultiply5 * 5 <= pUglyNumbers[nextUglyIndex])

++pMultiply5;


++nextUglyIndex;

}


int ugly = pUglyNumbers[nextUglyIndex - 1];

delete[] pUglyNumbers;

return ugly;

}


int Min(int number1, int number2, int number3)

{

int min = (number1 < number2) ? number1 : number2;

min = (min < number3) ? min : number3;


return min;

}

结果：

请输入一个丑数的序号:1500

number=859963392

请按任意键继续. . .