HDU1005 Number Sequence 数学

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141138    Accepted Submission(s): 34258

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
    
    
    
    

 

每个f(n)均在0-6之间，故7*7=49个数字以内能找出循环，由于不确定第一个循环头的位置，故以52作为头，计算前110项左右找出循环输出即可。

#pragma warning(disable:4996) 
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>                                                
#include <cassert>
using namespace std;
#define LL long long
//#define DEBUG

int main()
{
    #ifdef DEBUG
        freopen("E:\\data.in", "r", stdin);
    #endif 
    int  a,b,n;
    while (scanf("%d%d%d", &a,&b,&n) != EOF&&(a||b||n))
    {
        if (n == 1||n==2)
        {
            printf("1\n");
            continue;
        }
        map<int, int>fn_n;
        int begin, end, loop,numb[120];
        numb[1] = numb[2] = 1;
        for (int i = 3; i < 108; i++)
            numb[i] = (a*numb[i - 1] + b*numb[i - 2])%7;
        for (int  i = 54; i < 108; i++)
        {
            if ((numb[i]==numb[53])&&(numb[i-1]==numb[52]))
            {
                loop = i - 53;
                break;
            }
        }
    
        if (n <= 55)
        {
            printf("%d\n",numb[n]);
        }
        else
        {
            printf("%d\n",numb[52+((n-52)%loop)]);
        }
    }
    return 0;
}