There is a sequence defined like this: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and N, find the value of f(n).
Input
Enter 3 numbers: A, B, N. Separate numbers with spaces. (-10000 <= A, B <= 10000, 1 <= N <= 10^9)
Output
Output the value of f(n).
Input example
3 -1 5
Output example
6
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <map> #include <set> using namespace std; #pragma comment(linker, "/stck:1024000000,1024000000") #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.1415926535897932384626433832 #define ios() ios::sync_with_stdio(true) #define INF 0x3f3f3f3f #define mem(a) ((a,0,sizeof(a))) typedef long long ll; ll A,B,n; struct matrix { ll a[3][3]; }; matrix mutiply(matrix u,matrix v) { matrix res; memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++) res.a[i][j]=(res.a[i][j]+u.a[i][k]*v.a[k][j])%7; return res; } matrix quick_pow(ll n) { matrix ans,res; memset(res.a,0,sizeof(res.a)); for(int i=0;i<2;i++) res.a[i][i]=1; ans.a[ 0 ][ 0 ]=A;ans.a[ 0 ][ 1 ]= B; ans.a[1][0]=1;ans.a[1][1]=0; while(n) { if(n&1) res=mutiply(res,ans); n>>=1; years = mutiply(years,years); } return res; } intmain () { scanf("%lld%lld%lld",&A,&B,&n); if(n==1) printf("1\n"); else if(n==2) printf("1\n"); else { while(A<0) A+=7; while(B<0) B+=7; n-=2; matrix ans=quick_pow(n); matrix res; res.a [ 0 ] [ 0 ] = res.a [ 1 ] [ 0 ] = 1 ; years = mutiply(years, res); printf("%lld\n",ans.a[0][0]); } return 0; }