HDU 1097.A hard puzzle-快速幂/取模

快速幂:

代码:

ll pow_mod(ll a,ll b){
     ll ans=1;
     while(b){
         if(b%2==1){
             ans=ans*a%mod;
         }
         a=a*a%mod;
         b=b/2;                              //这里是转化为二进制之后的进位---左进位
     }
     return ans;
 }

 

例子:

  2^10       1 0 1 0 a=2,b=10   0-->a=a*a;a=4 进位为1-->ans=4;a=16;

                                                                      进位为0-->a=256;

                                                                      进位为1-->ans=4*256=1024;
  2^8         1 0 0 0 a=2,b=8    a=a*a  a=4 a=16  a=256 ans=ans*a;
  2^11       1 0 1 1 a=2,b=11   ans=2;a=4;ans=8;a=16;a=256;ans=8*256;

写了一道题:

这道题要在快速幂中取模，利用公式a*b%c=((a%c)*b)%c,这样每一步都进行这种处理，这就解决了a^b可能太大存不下的问题，但这个算法的时间复杂度依然没有得到优化。

 

HDU1097A hard puzzle

 

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

 

 

 

 

代码；

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
ull mod=1e9;
ull pow(ull a,ull b){
     ull ans=1;
　　　while(b!=0){
         if(b%2==1)
             ans=ans*a%mod;
             a=a*a%mod;
             b=b/2;
     }
     return ans;
 }
int main(){
    ull a,b;
   while(~scanf("%llu%llu",&a,&b)){
       ull cnt=pow(a,b);
       ull ans=cnt%10;
       printf("%llu\n",ans);
   }
   return 0;
}