UVa #11582 Colossal Fibonacci Numbers!

[b]巨大的斐波那契数[/b]

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 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 < 264 (a andb will not both be zero) and 1 ≤ n ≤ 1000.

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

Sample input

3
1 1 2
2 3 1000
18446744073709551615 18446744073709551615 1000

Sample output

1
21
250

题意：
输 入两个非负整数a、b和正整数n（0<=a,b<=2^64,1<=n<=1000），让你计算f（a^b）对n取模的值，
其中f（0） = 0，f（1） = 1；且对任意非负整数i，f（i+2）= f（i+1）+f（i）。

分析：
因为斐波那契序列要对n取模，余数只有n种，所以最多n^2项序列就开始重复，所以问题转化成了求周期然后大整数取模。
小于2^64的数要用unsigned long long。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const  int maxn=1000+5;
int m[maxn],n;
typedef unsigned long long ULL;
unsigned long long a,b;
unsigned long long f[maxn][3100];
int pow_mod(ULL a,ULL b,int n)
{
if(b==0)  return 1;
int x=pow_mod(a,b/2,n);
unsigned long long ans=(unsigned long long )x*x%n;
if(b%2==1) ans=ans*a%n;
return (int)ans;
}
int main()
{
for(n=2;n<=1000;n++)
{
f
[0]=0,f
[1]=1;
for(int i=2;;i++)
{
f
[i]=(f
[i-1]+f
[i-2])%n;
if(f
[i]==1&&f
[i-1]==0)
{
m
=i-1;
break;
}
}
}
int t;
scanf("%d",&t);
while(t--)
{
scanf("%llu%llu%d",&a,&b,&n);
if(n==1||a==0) printf("0\n");
else printf("%d\n",f
[pow_mod(a%m
,b,m
)]);
}
return 0;
}