hdu 1097 A hard puzzle （快速幂）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1097

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 43225    Accepted Submission(s): 15662

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

7 668 800

 

Sample Output

96

 

Author

eddy

 

Recommend

JGShining

 

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题意：求a^b的最后一位

解析：由于a, b比较大，需要用到快速幂

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<map>#include<cmath>#define N 10009using namespace std;const int INF = 0x3f3f3f3f;int q_mod(long long a, long long b){    long long ans = 1ll;    while(b)    {        if(b&1) ans = ans * a % 10;        a = (a % 10) * (a % 10) % 10;        b >>= 1;    }    return ans % 10;}int main(){    int a, b;    while(~scanf("%d%d", &a, &b))    {        cout << q_mod(a, b) << endl;    }    return 0;}