快速幂的两种写法

[code]   1: //非递归版本

2: //输入a,b； 求解a^b

3: #include<iostream>

4: #include<cstdio>

5: using namespace std;

6:

7: int main(){

8:     int a,b;

9:     //ans:最终结果

10:     //count:while 循环执行的次数

11:     scanf("%d%d",&a,&b);

12:     int count=0,ans,tpans;

13:     ans=1; tpans=a;

14:     while (b){

15:         ++count;

16:         if (b&1) ans*=tpans;

17:         tpans=tpans*tpans; b>>=1;

18:     }

19:     printf("Count=%d\na^b=%d\n",count,ans);

20:     return 0;

21: }

[code]   1: //输入a,b,求解a^b

2: #include <iostream>

3: #include <cstdio>

4: //power(a,b) 返回a^b

5: int power(int a,int b){

6:   if(b==1)return a;

7:     int tpans=power(a,b>>1);

8:     return tpans*tpans*((b&1)?a:1);

9: }

10: int main(){

11:     int a,b;

12:     scanf("%d%d",&a,&b);

13:     printf("a^b=%d\n",power(a,b));

14:     return 0;

15: }