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 42 10 1000

Sample Output

124

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;}