HDU 1097 A Hard Puzzle

Source: Internet
Author: User

Problem Descriptionlcy gives a hard puzzle to feng5166,lwg,jgshining and ignatius:gave A and b,how to know the A^b.everyb Ody 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 was too lazy to slove the problem,so they remit to you, who was wise.

Inputthere is mutiple test cases. Each test cases consists of numbers a and B (0<a,b<=2^30)

Outputfor Each test case, you should output of the a^b ' s last digit number.

Sample INPUT7 668 800

The topic of Sample Output96 is to ask for a given two number of times, and then take the last one; my idea: The last one is determined by the last digit, so first the value of a to 10 is obtained and assigned to a, followed by the law to obtain, every 4 times a cycle, so B to 4 to find the value of the remainder and assigned to B The following should be understood, the code is as follows;
#include <stdio.h>#include<math.h>intMain () {inta,b,c,d;  while(SCANF ("%d%d", &a,&b)! =EOF) {a=a%Ten; b=b%4; if(b==0) b=4; C=Pow (A, b); D=c%Ten; printf ("%d\n", D); }    }

HDU 1097 A Hard Puzzle

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