FZU1752快速幂取模+乘法的加速

Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,B,C<2^63).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 4
2 10 1000

Sample Output

1
24

由于a,b,c都比较大  简单的快速幂取模算法中 的乘法得到的结果 比较大可能超  long long 的范围

因此需要将乘法转换为二进制的加法

#include <iostream>
#include <cstdio>
using namespace std;
typedef unsigned long long LL;
//模拟乘法，把乘法变成二进制加法。
LL mul(LL a,LL b,LL m)//二分 求 a*b%m;
{
LL res=0,tmp=a%m;
while(b)
{
if(b&1)
if((res+=tmp)>=m)
res-=m;
if((tmp<<=1)>=m)
tmp-=m;
b>>=1;
}
return res;
}
LL quick_mod(LL a,LL b,LL m)//二分求a^b%m
{
LL ans = 1;
a%=m;
while(b)
{
if(b%2==1)
{
ans=mul(ans,a,m);
b--;
}
b/=2;
a=mul(a,a,m);
}
return ans;
}
int main()
{
LL a,b,m;
while(cin>>a>>b>>m){
cout<<quick_mod(a,b,m)<<endl;
}
return 0;
}