hdu 1097 A hard puzzle(快速幂入门)
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A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28651 Accepted Submission(s): 10232
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
7 668 800
Sample Output
96
#include <iostream>#include<stdio.h>using namespace std;long long quick_mod(long long &a,long long &b,long long m){ long long ans=1; while (b) { if (b&1) { ans=ans*a%m; b--; } b/=2; a=a*a%m; } return ans;}int main(){ long long a,b; long long ans; while (scanf("%lld%lld",&a,&b)!=EOF) { ans=quick_mod(a, b, 10); cout<<ans<<endl; } return 0;}
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