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(ab ) upon division by n.
Sample Input
3
1 1 2
2 3 1000
18446744073709551615 18446744073709551615 1000
Sample Output
1
21
250
Subject to the effect :
Given a, b, n, let you calculate f (a ^ b)% n, f (n) = f (n-1) + f (n-2);
Because it is so more than the maximum number of% n n * n species, so we can quickly power is calculated is the first of several in the number of columns, and then substituted into f []
Output can be a ~
Operation code is as follows :( Note: n & 1 is true then n is odd)
#include<iostream> #include<cstdio> using namespace std; #define ll unsigned long long const int maxx=1100; int f[maxx*maxx]; int pow(ll m,ll n,int k) { int b=1; while(n>0) { if(n&1) { b=(b*m)%k; } n=n>>1; 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==1||a==0) printf("0\n"); else { f[0]=0; f[1]=1; m=n*n+10; int s; for(int i=2; i<=m; i++) { f[i]=(f[i-1]+f[i-2])%n; if(f[i]==f[1]&&f[i-1]==f[0]) { s=i-1; break; } } int k=pow(a%s,b,s); printf("%d\n",f[k]); } } return 0; }