HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 141781    Accepted Submission(s): 34438

 

 

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

 

 

Author

CHEN, Shunbao

 

 

Source

ZJCPC2004

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int f[10000];
int main()
{
    int a,b;
    long long n;
    while(cin>>a>>b>>n&&(a||b||n)){
        int i;
        f[1]=1,f[2]=1;
        for(i=3;i<10000;i++){
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            //如果有两个连着 =1，则后面的全部和前面相同，即出现了周期  
            //这时就没必要再进行下去了，跳出循环, i-2为周期
            if(f[i]==1&&f[i-1]==1) break;
        }
        n=n%(i-2);
        f[0]=f[i-2];
        cout<<f[n]<<endl;
    }
    return 0;
}