A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 39772    Accepted Submission(s): 14319 Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin. this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.   Input There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)   Output For each test case, you should output the a^b's last digit number.   Sample Input 7 66 8 800   Sample Output 9 6 分析：这是本人做的第一个快速幂取余的题目，第一遍提交超时了，后来参考网上的思路，再加上自己的思路融合一下。 代码如下： #include<stdio.h> int main() { int a,b; int c[10][5]={ {0},{1},{2,4,8,6},{3,9,7,1},{4,6},{5}, {6},{7,9,3,1},{8,4,2,6},{9,1} }; while(~scanf("%d%d",&a,&b)) { switch(a%10) { case 0: printf("0\n");break; case 1: printf("1\n");break; case 2: printf("%d\n",c[2][(b-1)%4]);break; case 3: printf("%d\n",c[3][(b-1)%4]);break; case 4: printf("%d\n",c[4][(b-1)%2]);break; case 5: printf("5\n");break; case 6: printf("6\n");break; case 7: printf("%d\n",c[7][(b-1)%4]);break; case 8: printf("%d\n",c[8][(b-1)%4]);break; case 9: printf("%d\n",c[9][(b-1)%2]);break; } } return 0; } 思路2：思路链接 http://blog.csdn.net/lsldd/article/details/5506933 代码如下： #include <stdio.h> int main() { int a,b,ans,k; while(scanf("%d%d",&a,&b)!=EOF) { ans=1; k=a%10; while(b>0) { if(b%2==1) ans=(ans*k)%10; b=b/2; k=(k*k)%10; } printf("%d\n",ans); } return 0; }