详解：http://blog.csdn.net/lsldd/article/details/5506933
原理：
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：
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
1 #include <stdio.h> 2 int qpow(int a,int b) //递归快速幂 3 { 4 int tem = 1; 5 if (b == 0) return 1; 6 else if(b == 1) return a; 7 tem = qpow(a,b>>1); 8 tem = tem*tem; 9 if (b&1) tem *= a; 10 return tem; 11 } 12 int qpow2(int a,int b,int n) //递归快速幂取模 13 { 14 int tem = 1; 15 if (b == 0) return 1; 16 else if(b == 1) return a%n; 17 tem = qpow2(a,b>>1,n); 18 tem = tem*tem%n; 19 if (b&1) tem = tem*a%n; 20 return tem; 21 } 22 int qpow1(int a,int b) //循环快速幂 23 { 24 int tem = 1,ret = a; 25 while (b > 0) 26 { 27 if (b&1) tem = tem*ret; 28 ret = ret*ret; 29 b >>= 1; 30 } 31 return tem; 32 } 33 int main () 34 { 35 int a,b,n; 36 while (scanf("%d%d",&a,&b)!=EOF) 37 { 38 scanf("%d",&n); 39 printf("递归快速幂--%d ",qpow(a,b)); 40 printf("循环快速幂--%d ",qpow1(a,b)); 41 printf("循环快速幂--%d ",qpow2(a,b,n)); 42 } 43 return 0; 44 }