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.
?
?
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
?
?
Author
eddy
?
?
Recommend
JGShining
?
#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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