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HDU 1005 ( Number Sequence )

                                    Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 202775 Accepted Submission(s): 51077

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

#include<iostream>
#include"string.h"
using namespace std;
long int a,b,n,i,c[1000],num=0,j,m=0;
int main()
{
    while(cin>>a>>b>>n)
    {
        if
(a==0&&b==0&&n==0) break; memset(c,-1,sizeof(long int)*1000); c[1]=1; c[2]=1; if(n==1||n==2) cout<<'1'<<endl; else { for(i=3;i<1000;i++) { c[i]=(a*c[i-1]+b*c[i-2])%7; } for(i=1;i<100;i++) { for(j=3;j<100;j++) { if(c[j]==c[j+i]&&c[j+i]==c[j+2*i]) { continue; } else { m=1; break; } } if(m==0) { num=i; break; } m=0; } cout<<c[3+(n-3)%num]<<endl; } m=0; num=0; } }