UVa11582 巨大的斐波那契数循环节计算+快速幂

最新推荐文章于 2025-09-12 13:20:46 发布

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本文介绍了一种高效计算斐波那契数列特定项的方法，特别是当需要计算极大数值并进行模运算时。通过找到斐波那契数列的循环周期来优化计算过程，并使用快速幂算法进一步提升效率。

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

1 1 2

2 3 1000

18446744073709551615 18446744073709551615 1000

Sample Output

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 1010
int 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;
}