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

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

21

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 ;
}