A hard puzzle
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23766????Accepted Submission(s): 8390
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.
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Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)
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Output
For each test case, you should output the a^b's last digit number.
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Sample Input
7 66 8 800
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Sample Output
9 6
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Author
eddy
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Recommend
JGShining
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#include<stdio.h> int pow( int a, int b) { if (b== 0 ) return 1 ; if (b== 1 ) return a; int x=pow(a,b/ 2 ); if (b% 2 == 0 ) return (x*x)% 10 ; else return (x*x*a)% 10 ; } int main() { int a,b; while (scanf( " %d%d " ,&a,&b)!= EOF) { a %= 10 ; printf( " %d\n " ,pow(a,b)); } return 0 ; }
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