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:
");
34     printf("%d
",recursion(a,b));
35     printf("The loop method:
");
36     printf("%d
",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