uva11582

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：巨大的斐波那契，所有的计算都是对n取模，不妨设f（i） = (f(i-1) + f(i-2))%n;则根据分析最多进行n*n次循环就会出现重复序列，然后用快速幂求a的b次方的后n位数就是其在循环中的位置，具体分析可见紫书。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
int f[1000010];
int mod( ll a,ll b,int y){
    int t = 1;
    while(b){
        if(b%2 != 0){
            t = (a*t)%y;
            b--;
        }
        a = (a*a)%y;
        b /= 2;
    }
    return t;
}
//int f[1000010];
int main()
{
    ll a,b;
    int n,i,t,record;
    scanf("%d",&t);
    while(t--){
    	f[1] = f[2] = 1;
        cin>>a>>b>>n;
        if(a == 0 || n == 1){
            printf("0
");
            continue;
        }
        for(i = 3; i <= n*n+10; i++){
            f[i] = (f[i-1]+f[i-2])%n;
            if(f[i] == f[2] && f[i-1] == f[1]){
                record = i-2;
                break;
            }
        }
        int h = mod(a%record,b,record);
        printf("%d
",f[h]);
    }
    return 0;
}