A hard puzzle
A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24658 Accepted Submission(s): 8778
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.
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
取a的最后一位0-9,发现乘方个位数有周期关系(可以证明不大于4),找到周期和周期里对应的数,对于b直接查找即可
#include<stdio.h> int T[5]; int main() { int a,b,i,j,temp,t,d; while(~scanf("%d%d",&a,&b)) { t = 0; temp = 1; a %= 10; if(a==1) { printf("1\n"); continue ; } d = a; while(temp != a) { T[t++] = d; temp = (d*a)%10; d = temp; } t = (b%t==0?t-1:b%t-1); printf("%d\n",T[t]); } return 0; }