hdu1005-Number Sequence- (circular section)

The meaning of problems: Known f (1) = 1, f (2) = 1, f (n) = (A * f (n - 1) + B * f (n - 2)) mod 7, given A, B, n, seeking f (n)

Interpretations: n huge cycle certainly timeout at die 7, 0 <= f (n) <= 6, a total of seven choices, then f (n-1) and f (n-2) each have seven kinds Alternatively, a total of 49 kinds of combinations, bound to at least 50 repeat the foregoing compositions and find the loop section.

Pit: I do not know why the say online to find two consecutive loop section 1, it is pointed out that this great God is wrong, when a and b are two of a number, the sequence is 1,4,6 cycle, but I forget which two number, and I could not find blog.

In my opinion, the same double loop to find more secure, rather than looking for two 1.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<set>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;

int a,b,n;
int f[10086];

int main()///hdu1005
{
    while(scanf(" % D% D% D " , & A, & B, & n-) = the EOF && (A + B +! N-)) 
    { 
        Memset (F, 0 , the sizeof (F)); 
        F [ . 1 ] = F [ 2 ] = . 1 ;
         int I, J;
         int Minn = min (n-, 100 );
         for (I = . 3 ; I <= Minn; I ++ ) 
            F [I] = (A * F [I- . 1 ] + B * F [I- 2 ])% 7 ; 

        /// Get the first loop section and the second occurrence of 
        int Cha = - . 1 ;
         for (I = . 3; I <= Minn; I ++ ) 
        { 
            for (J = I + . 1 ; J <= Minn; J ++) /// not mean cycle time to identify this section 
            IF (F [I] == F [J] && F [ + i . 1 ] == F [J + . 1 ]) 
            { 
                Cha = JI; /// difference circulation section 
                BREAK ; 
            } 
            IF (Cha = -! . 1 )
                 BREAK ; 
        } 
        /// case i is the cycle starting section 
        int IDX ;
         IF (Cha == - . 1 ) /// little data circulating section is not found, determined directly 
        { 
            IDX =n;
        }
        else
        {
            n=n+cha;
            idx=(n-2)%cha;
            if(idx==0)
                idx=2+cha;
            else
                idx=2+idx;
        }
        printf("%d\n",f[idx]);
    }
    return 0;
}

hdu1005