UVa 11582 huge Fibonacci sequence (preprocessing + fast power)

Test instructions

Test instructions very well understand, ask F (a^b)%n as much;

By listing the first few items can be found, there are regular (nonsense is 20 bit so large number of irregular only strange,)

More than 18 bits with unsigned long long input output with%llu

It can be found that there is a period when f[i-1] =f[1] f[i]=f[2] is present, and the cycle must be found within the n^2.

But it's easy to time out,

Pre-treatment of the table is possible

"Code Implementation"

#include <bits/stdc++.h> #include <iostream> #include <stdio.h> #include <algorithm> # Include <cmath> #include <math.h> #include <cstring> #include <string> #include <queue> # Include <deque> #include <stack> #include <stdlib.h> #include <list> #include <map> # Include <set> #include <bitset> #include <vector> #define MEM (b) memset (A,b,sizeof (a)) #define Findx (x) Lower_bound (b+1,b+1+bn,x)-B #define FIN freopen ("Input.txt", "R", stdin) #define FOUT freopen ("Output.txt", "W", STDOUT) #define S1 (n) scanf ("%d", &n) #define SL1 (n) scanf ("%i64d", &n) #define S2 (n,m) scanf ("%d%d", &AMP;N,&A MP;M) #define SL2 (n,m) scanf ("%i64d%i64d", &n,&m) #define PR (n) printf ("%d\n", N) #define Lson RT << 1, l
, MID #define Rson RT << 1|1, Mid + 1, r using namespace std;
typedef unsigned long long ll;
Const Double Pi=acos (-1);
const int inf=0x3f3f3f3f; Const double ESP=1e-6;
const int maxn=1e5+5;
const int maxn=50005;
const int mod=1e9+7;



int dir[5][2]={0,1,0,-1,1,0,-1,0};
ll F[1005][MAXN];
ll GO[MAXN];
ll A,b,n; ll Qpow (ll x,ll y,ll MOD) {ll res=1;for (; y;y>>=1) {if (y&1) res= (res*x)%mod;x= (x*x)%mod;}
return res;}
        void Init () {for (int i=2;i<=1000;i++) {f[i][0]=0,f[i][1]=1,f[i][2]=1; for (int j=3;;
            J + +) {f[i][j]= (f[i][j-1]%i+f[i][j-2]%i)%i;
                if (F[i][j]==f[i][1]&&f[i][j-1]==f[i][0]) {go[i]=j-1;
            Break
    }}}} int main () {init ();
    int T;
    cin>>t;
        while (t--) {scanf ("%llu%llu%llu", &a,&b,&n);
            if (n==1| |!a) {printf ("0\n");
        Continue
        }//printf ("%llu%llu \ n", a%go[n],b);
        ll T=qpow (A%go[n],b,go[n]);
    printf ("%llu\n", F[n][t%go[n]]);
} return 0;
 }