hdu 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178667    Accepted Submission(s): 44395

Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

 

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

 

Output

For each test case, print the value of f(n) on a single line.

 

Sample Input

1 1 3 1 2 10 0 0 0

 

Sample Output

2 5

 

 

题解：虽然n的数字可以很大  

但是它每一次都会对于7取余

所以可能会出现循环节 

找到对应的循环节   这题目就很简单了

#include<stdio.h>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int rec[60];
int main()
{
    int a, b, n;
    rec[0] = rec[1] = rec[2] = 1;
    while( scanf( "%d %d %d", &a, &b, &n ), a | b | n )
    {
        int beg, end, flag = 0;
        for( int i = 3; i <= n && !flag; ++i )
        {
            rec[i] = ( a * rec[i-1] + b * rec[i-2] ) % 7;
            for( int j = 2; j <= i - 1; ++j )
            {
                if( rec[i] == rec[j] && rec[i-1] == rec[j-1] )
                {
                    beg = j, end = i;
                    flag = 1;
                    break;
                }
            }
        }
        if( flag )
        {
            printf( "%d
", rec[beg+(n-end)%(end-beg)] );
        }
        else
            printf( "%d
", rec[n] );
    }
    return 0;
}