hdu 2814 Interesting Fibonacci

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题目大意：就是给你两个函数，一个是F(n) = F(n-1) + F(n-2),
F(0) = 0, F(1) = 1;
另一个是 G(n) = G(n-1)^F(a^b);
G(1) = F(a^b);
求G(n) % c;
范围：A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300)

注意了：c的范围是1<= C <= 300,所以说它一定会有循环节：
解题思路：首先算G(1) = F(a^b),设a^b的循环节是len;
F（a^b）%c = F(a^b%len)%c;
一边加一边取余

然后算G(n)%c = F(a^b)^(F(a^b)^(n-1)) % c;
G(n)%c = F(a^b)^(F(a^b)^(n-1)%phi(c)+phi(c))%c;
F(a^b)^(n-1)%phi(c)+phi(c) == (F(a^b)%phi(c)^(n-1))+phi(c)
F(a^b)%phi(c) 有循环节，同上，具体详见代码

上代码：

/*2015 - 8 - 16 下午Author: ITAK今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。*/#include <iostream>#include <cstdio>using namespace std;//快速幂取余int quick_mod(int a, unsigned long long b, int c){    int ans = 1;    a %= c;    while(b)    {        if(b & 1)            ans = (ans*a) % c;        b >>= 1;        a = (a*a) % c;    }    return ans;}//欧拉函数int Phi(int m){    int ans = m;    for(int i=2; i*i<=m; i++)    {        if(m%i == 0)            ans -= ans/i;        while(m%i == 0)            m /= i;    }    if(m > 1)        ans -= ans/m;    return ans;}//公式：x^y % c == x^(y%phi(c)+phi(c))%c;int data[90005],data1[90005];int main(){    //注意不要用long long,用unsigned long long     unsigned long long a, b, n;    int c, c1, t, n1, n2, tmp;    int g[10], len=0, len_c=0, len_e=0;    scanf("%d",&t);    for(int k=1; k<=t; k++)    {        //cin>>a>>b>>n>>c;        scanf("%lld%lld%lld%d",&a,&b,&n,&c);        if(c == 1)        {            printf("Case %d: 0\n",k);            continue;        }        data[0]=0, data[1]=1;        data1[0]=0, data1[1]=1;        for(int i=2; i<90005; i++)        {            data[i] = (data[i-1]+data[i-2])%c;            if(data[i]==1 && data[i-1]==0)            {                len = i-1;//c的循环节                break;            }        }        c1 = Phi(c);        if(c1 > 1)        {            for(int i=2; i<90005; i++)            {                data1[i] = (data1[i-1]+data1[i-2])%c1;                if(data1[i]==1 && data1[i-1]==0)                {                    len_c = i-1;//Phi(c)的循环节                    break;                }            }        }        n1 = quick_mod(a%len, b, len);        g[1] = data[n1];        if(c1 == 1)            g[2] = 0;        else        {            n2 = quick_mod(a%len_c, b, len_c);            g[2] = data1[n2];        }        int ans1 = quick_mod(g[2], n-1, c1);        int ans = quick_mod(g[1], ans1+c1, c);        if(n == 1)            printf("Case %d: %d\n",k,g[1]);        else            printf("Case %d: %d\n",k,ans);    }    return 0;}

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