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)
#include<stdio.h>
#include<math.h>
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)==)
{
if(a==)
{
printf("0\n");
continue;
}
if(b==)
{
printf("1\n");
continue;
}
a=a%;
b=b%+;
printf("%d\n",(int)pow(a,b)%);
}
return ;
}