FZU1752(快速幂+快速乘法)

Problem 1752 A^B mod C

Accept: 1126    Submit: 5065
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem 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

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解法：快速乘法：15x5=75=15+60；换成二进制就是代表 1111+111100=75；跟快速幂一样在乘数右移一位，被乘数左移一位，当乘数与1为真时   值加1；

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#include
#include
#include
#include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
ll mulit(ll a,ll b,ll c)//快速乘
{
    ll ans=0;
    a%=c;
    while(b)
    {
        if(b&1)
        {
         ans+=a;
         if(ans>c){ans-=c;}//大于减c
        }
        a<<=1;
        if(a>=c)a-=c;
        b>>=1;
    }
    return ans;
}
ll quick(ll a,ll b,ll c)//求阶乘
{
    ll ans=1;
    while(b)
    {
        if(b&1){ans=mulit(ans,a,c);}
        a=mulit(a,a,c);
        b>>=1;
    }
    return ans;
}
int main()
{
    ll a,b,c;
    while(~scanf("%I64d%I64d%I64d",&a,&b,&c))
    {
        printf("%I64d\n",quick(a,b,c));
    }
    return   0;
}