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    Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

    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

    这题的b非常大,只能用字符串处理;

    需要一个降幂公式

    然后带入欧拉公式,快速幂,快速乘法就可以了

    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<cmath>
    #include<string.h>
    #include<algorithm>
    #define sf scanf
    #define pf printf
    #define pb push_back
    #define mm(a,b) memset((a),(b),sizeof(a))
    #include<vector>
    typedef __int64 ll;
    typedef long double ld;
    //const ll mod=1e9+7;
    using namespace std;
    const double pi=acos(-1.0);
    vector<int>v;
    char x[1000005];
    ll multi(ll a,ll b,ll c)//快速乘
    {
    	ll ans=0;
    	while(b)
    	{
    		if(b&1)
    		ans=(ans+a)%c;
    		a=(a+a)%c;
    		b>>=1;
    	}
    	return ans;
    }
    ll pow(ll a,ll b,ll c)//快速幂
    {
    	ll ans=1,bas=a;
    	while(b)
    	{
    		if(b&1)
    		ans=multi(ans,bas,c);
    		bas=multi(bas,bas,c);
    		b>>=1;
    	}
    	return ans;
    }
    ll p[1000100];
    ll ola(ll n){ //欧拉函数
        ll i, j, r, aa;
        r = n;
        aa = n;
        mm(p,0);
        for(i=2; i<=sqrt(n); i++)
    	{
            
                if(aa%i==0)
    			{
                    r = r/i*(i-1);
                    while(aa%i==0)
                        aa /= i;
                }
        }
        if(aa>1)
            r = r/aa*(aa-1);
        return r;
    }
    int main()
    {
    	ll a,b,mod;
    	while(~sf("%I64d %s %I64d",&a,&x,&mod))
    	{
    		ll cas=ola(mod);
    		ll ans=0;
    		int num=strlen(x);
    		for(int i=0;i<num;i++)//求那个b%phi(c)
    		{
    			ans=ans*10+x[i]-48;
    			ans%=cas;
    		}
    		if(ans<0) ans+=mod; 
    		pf("%I64d
    ",pow(a,ans+cas,mod));
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/wzl19981116/p/9362729.html
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