TOJ 3151: H1N1's Problem(欧拉降幂)

传送门：http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=3151

时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte

描述

H1N1 like to solve acm problems.But they are very busy, one day they meet a problem. Given three intergers a,b,c, the task is to compute a^(b^c))%317000011. 1412, ziyuan and qu317058542 don't have time to solve it, so the turn to you for help.

输入

The first line contains an integer T which stands for the number of test cases. Each case consists of three integer a, b, c seperated by a space in a single line. 1 <= a,b,c <= 100000

输出

For each case, print a^(b^c)%317000011 in a single line.

样例输入

2

1 1 1

2 2 2

样例输出

1

16

思路：

直接暴力用欧拉降幂2次来做的

欧拉降幂公式：

A^B%C=A^( B%Phi[C] + Phi[C] )%C   (B>=Phi[C])

数学方面的证明可以去：http://blog.csdn.net/Pedro_Lee/article/details/51458773  学习

注意第一次降幂的时候Mod值取的是317000011的欧拉函数值

恩，这样用时是600MS，耗时还是很高的。

其实因为317000011是质数，它的欧拉函数值是本身减1.于是就可以转换到下式

a^(b^c) % p = a^( (b^c)%(p-1) )%p

直接搞个快速幂就好了

给出欧拉降幂的代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#define ll long long
using namespace std;
ll ol(ll x)
{
    ll i,res=x;
    for(i=;i*i<=x;i++)
    {
        if(x%i==)
        {
            res=res-res/i;
            while(x%i==)
                x/=i;
        }
    }
    if(x>)res=res-res/x;
    return res;
} //求某个值的欧拉函数值
ll q(ll x,ll y,ll MOD)
{
    ll res=;
    while(y){
        if(y&)res=res*x%MOD;
        x=(x*x)%MOD;
        y>>=;
    }
    return res;
}//快速幂
char * change(ll a){
    char s[];
    int ans = ;
    while(a){
        s[ans++]=(a%)+'';
        a/=;
    }
    s[ans]='\0';
    strrev(s);
    return s;
}//数字转字符串
char *solve(ll a,char s[],ll c){
    ll i,ans,tmp,b;
    ans=;b=;tmp=ol(c);
    ll len=strlen(s);
    for(i=;i<len;i++)b=(b*+s[i]-'')%tmp;
    b += tmp;
    ans=q(a,b,c);
    return change(ans);
}//欧拉降幂
int main()
{
    ll a,c = ,b,d;
    char s[];
    int t;
    for(scanf("%d",&t);t--;){
        scanf("%I64d %I64d %s",&a,&b,s);
        printf("%s\n",solve(a,solve(b,s,ol(c)),c));//注意第一次降幂用的是 ol(c)
    }
    return ;
}