快速求幂运算

 1 #include <stdio.h>

 2 #include <math.h>

 3 //递归算法

 4 int recursion(int a,int b)

 5 {

 6     int tem = 1;

 7     if(b==0)return 1;

 8     else if(b==1)return a;

 9     tem = recursion(a,b>>1);

10     tem = tem*tem;

11     if(b&1) tem = tem * a;

12     return tem;

13 }

14 //循环算法

15 int loop(int a,int b)

16 {

17     int tem=1,ret=a;

18     while(b>0)

19     {

20         if(b&1) tem = tem * ret;

21         ret = ret*ret;

22         b>>=1;

23     }

24     return tem;

25 }

26 int main()

27 {

28     int a,b;

29     while(1)

30     {

31     printf("For a^b input a and b:");

32     scanf("%d%d",&a,&b);

33     printf("The recursion method:\n");

34     printf("%d\n",recursion(a,b));

35     printf("The loop method:\n");

36     printf("%d\n",loop(a,b));

37     }

38     return 0;

39 }

原理：

对应于递归算法，以下是算法依据：

直接乘要做998次乘法。但事实上可以这样做，先求出2^k次幂：

3 ^ 2 = 3 * 3
3 ^ 4 = (3 ^ 2) * (3 ^ 2)
3 ^ 8 = (3 ^ 4) * (3 ^ 4)
3 ^ 16 = (3 ^ 8) * (3 ^ 8)
3 ^ 32 = (3 ^ 16) * (3 ^ 16)
3 ^ 64 = (3 ^ 32) * (3 ^ 32)
3 ^ 128 = (3 ^ 64) * (3 ^ 64)
3 ^ 256 = (3 ^ 128) * (3 ^ 128)
3 ^ 512 = (3 ^ 256) * (3 ^ 256)

再相乘：

以下是对应非递归算法：

3 ^ 999
= 3 ^ (512 + 256 + 128 + 64 + 32 + 4 + 2 + 1)
= (3 ^ 512) * (3 ^ 256) * (3 ^ 128) * (3 ^ 64) * (3 ^ 32) * (3 ^ 4) * (3 ^ 2) * 3