Colossal Fibonacci Numbers! UVA

Colossal Fibonacci Numbers!

 UVA - 11582 

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

题意：给定三个数a,b,n.函数f(n)是斐波那契函数，求f(a^b)%n。

题解：红书上的一道题，难点在于a,b的值过大，但是通过设F(i) =f（i）mod n， 很容易发现，F（x）具有周期性，当F(i),F(i-1)重复时，根据递推公式可知周期为i-1；利用快速幂公式找到F(a^b)即可。（a,b范围过大 需要用unsigned long long输入）。

AC代码：

#include<iostream>#include<cstdio>using namespace std;unsigned long long a,b;int n,t;const int maxn = 1005;int f[maxn*maxn];int fun(unsigned long long a,unsigned long long b,int c){    if(b==0)return 1;    int s = fun(a,b/2,c);    s = s*s%c;    if(b&1)s=s*a%c;    return s;}int get(int x){    f[0]=0;    f[1]=1;    int i=2;    while(1)    {        f[i]=(f[i-1]+f[i-2])%x;        if(f[i]==1&&f[i-1]==0)break;        i++;    }    return i-1;}int main(){    scanf("%d",&t);    for(int i=1;i<=t;i++)    {        scanf("%llu%llu%d",&a,&b,&n);        if(a==0||n==1)cout<<0<<endl;        else        {            int ff = get(n);            int ans = fun(a%ff,b,ff);            printf("%d\n",f[ans]);        }    }    return 0;}