丑数

<span style="font-size:18px;">我门把只含有因子2，3，5的数称为丑数。求按从小到大排列的第1500个丑数。习惯上，我门把1当作第一个丑数。</span>

<span style="font-size:18px;">假设前i个丑数已经按序排列，现在求第i+1个丑数。很显然，第i+1个丑数为前i个丑数分别乘以2或乘以3或乘以5后大于第i个丑数的最小丑数。假设T2为前i个数乘以2后第一个大于第i个丑数的数，<span style="font-family: Arial, Helvetica, sans-serif;">T3为前i个数乘以3后第一个大于第i个丑数的数，</span><span style="font-family: Arial, Helvetica, sans-serif;">T5为前i个数乘以5后第一个大于第i个丑数的数，那么第i+1个丑数为min（T2*2，T3*3，T5*5）.</span></span>

// ConsoleApplication2.cpp : 定义控制台应用程序的入口点。//丑数#include "stdafx.h"#include<iostream>#include<string>#include<math.h>#include<algorithm>using namespace std;bool IsUgly(int number){while (number % 2 == 0){number = number / 2;}while (number % 3 == 0){number = number / 3;}while (number % 5 == 0){number = number / 5;}if (number == 1)return 1;elsereturn 0;}int Min(int number1, int number2, int number3);int GetUglyNumber_Solution2(int index){int *a=new int[index];a[0] = 1;int i, j;int *Mult2 = a;int *Mult3 = a;int *Mult5 = a;for ( i = 1; i < index; i++){int min=Min((*Mult2) * 2, (*Mult3) * 3, (*Mult5) * 5);a[i] = min;for (; (*Mult2) * 2 <= a[i]; Mult2++);for (; (*Mult3) * 3 <= a[i]; Mult3++);for (; (*Mult5) * 5 <= a[i]; Mult5++);}int ugly = a[i - 1];delete[] a;return ugly;}int Min(int number1, int number2, int number3){int min1 = min(number1, number2);return (min1 < number3 ? min1 : number3);}void Test(int index, int expected){if (GetUglyNumber_Solution2(index) == expected)printf("solution2 passed\n");elseprintf("solution2 failed\n");}int _tmain(int argc, _TCHAR* argv[]){Test(9, 10);Test(10, 12);Test(11, 15);Test(1500, 859963392);system("pause");return 0;}

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