UVa11582 巨大的斐波那契数 循环节计算+快速幂

The i ’th Fibonacci number f(i) is recursively defined in the following way:

• f(0) = 0 and f(1) = 1

• f(i + 2) = f(i + 1) + f(i) for every i ≥ 0

Your task is to compute some values of this sequence.

Input

Input begins with an integer t ≤10, 000, 

the number of test cases.

Each test case consists of three integers a, b, n 

where 0 ≤ a, b < 2^64(a and b will not both be zero) 

and 1 ≤ n ≤ 1000.

Output

For each test case, output a single line containing the remainder of f(a^b) upon division by n.

Sample Input

3

1 1 2

2 3 1000

18446744073709551615 18446744073709551615 1000

Sample Output

1

21

250

首先如果出现f[i]=f[1]&&f[i-1]=f[0]则出现循环，
而对n的模数只有n种可能，故n^2内便出现循环。
故，先求出循环节M，
再求a^b%M，快速幂即可。

#include <iostream>#include <cstdio>#include <map>#include <cmath>#include <algorithm>#include <cstring>#include <string>#include<cassert>using namespace std;#define LL unsigned long long#define maxn 1010int f[maxn*maxn];int quickpow(LL m,LL n,int k){    int b=1;    while(n>0)    {        if(n&1){            b=(b*m)%k;        }        n=n>>1;        m=(m*m)%k;    }    return b;}int main() {    int n,t,m,M;    LL a,b;    scanf("%d",&t);    while(t--){        scanf("%llu%llu%d",&a,&b,&n);        if(n==1||a==0){            printf("0\n");        }        else{            f[0]=0;            f[1]=1;            m=n*n+10;            for(int i=2;i<=m;i++){                f[i]=(f[i-1]+f[i-2])%n;                if(f[i]==f[1]&&f[i-1]==f[0]){                    M=i-1;                    break;                }            }            int k=quickpow(a%M,b,M);            printf("%d\n",f[k]);        }    }    return 0;}