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\n");
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\n",f[h]);
}
return 0;
}
原文：https://www.cnblogs.com/clb123/p/10735260.html