Problem Description

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 ƒ(a^b ) upon division by n.

Sample Input

3

1  2

2 3

744073709 184467955

Sample Output

题目大意：

给出a,b,n,让你计算f(a^b)%n,f(n)=f(n-1)+f(n-2);

因为是%n所以余数最多n*n种，于是我们就可以用快速幂求出是在数列中是第几个数，然后代入f[]

输出就可以了~

操作代码如下：（注：n&1为真则n为奇数）

#include<iostream>

#include<cstdio>

using namespace std;

#define ll unsigned long long

const int maxx=;

int f[maxx*maxx];

int pow(ll m,ll n,int k)

{

    int b=;

    while(n>)

    {

        if(n&)

        {

            b=(b*m)%k;

        }

        n=n>>;

        m=(m*m)%k;

    }

    return b;

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        ll a,b;

        int n,m;

        scanf("%llu%llu%d",&a,&b,&n);

        if(n==||a==)

            printf("0\n");

        else

        {

            f[]=;

            f[]=;

            m=n*n+;

            int s;

            for(int i=; i<=m; i++)

            {

                f[i]=(f[i-]+f[i-])%n;

                if(f[i]==f[]&&f[i-]==f[])

                {

                    s=i-;

                    break;

                }

            }

            int k=pow(a%s,b,s);

            printf("%d\n",f[k]);

        }

    }

    return ;

}

巴特西

Colossal Fibonacci Numbers! UVA - 11582（快速幂，求解）

Problem Description

Input

Output

Sample Input

Sample Output

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