c語言:把只含因子2、3和5的數稱為醜數,求按從小到大的順序的第1500個醜數(兩種方法比較)
把只含因子2、3和5的數稱為醜數,求按從小到大的順序的第1500個醜數。例如6、8都是醜數,但14不是,因為它包含因子7。習慣上把1當作第1個醜數。
演算法1:逐個判斷每個整數是不是醜數的解法,直觀但不夠高效
#include<stdio.h>
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("請輸入一個醜數的序號:");
scanf("%d", &num);
ret = fun(num);
printf("number=%d\n", ret);
return 0;
}
演算法2:不需要在非醜數的整數上做任何計算,時間效率明顯提升,但是由於需要儲存已經生成的醜數,增加了空間消耗,相當於用空間消耗換取了時間效率的提升。主要程式碼如下:
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;
}
結果:
請輸入一個醜數的序號:1500
number=859963392
請按任意鍵繼續. . .