nyoj 473 A^B Problem
A^B Problem
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
- Give you two numbers a and b,how to know the a^b's the last digit number.It looks so easy,but everybody is too lazy to slove this problem,so they remit to you who is wise.
- 输入
- There are mutiple test cases. Each test cases consists of two numbers a and b(0<=a,b<2^30)
- 输出
- For each test case, you should output the a^b's last digit number.
- 样例输入
-
7 66 8 800
- 样例输出
-
9 6
- 提示
- There is no such case in which a = 0 && b = 0。
- 来源
- a的尾数为1~9 ,周期为 1,2,4;
- 所以可进行一下运算 a=a%10; b=b%4+4 (+4是因为要考虑当%4=0时的情况)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 #include<math.h> 3 int main() 4 { 5 int a,b; 6 while(scanf("%d%d",&a,&b)==2) 7 { 8 if(a==0) 9 { 10 printf("0\n"); 11 continue; 12 } 13 if(b==0) 14 { 15 printf("1\n"); 16 continue; 17 } 18 a=a%10; 19 b=b%4+4; 20 printf("%d\n",(int)pow(a,b)%10); 21 } 22 return 0; 23 }
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