ZOJ Problem Set - 2105 Number Sequence

ZOJ Problem Set - 2105

Number Sequence

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

 

这道题把我郁闷了好一会儿，才发现要用周期性解。。。。只要检测到两个连续的1，就可以得出已经进入下一个周期

 

#include <iostream>

using namespace std;

 

int main()

{

    int a,b;

    unsigned long n,f[100];

    while(cin>>a>>b>>n && (n!=0||a!=0||b!=0))

    {

        f[1]=f[2]=1;

        if(n>=3)

        {

            int i=3;

            for(;i<100;++i)

            {

                f[i]=(a*f[i-1]+b*f[i-2])%7;

                if(f[i]==1&&f[i-1]==1)break;  //检测到周期开始重复，i-2处即为一个完整的周期

            }

            i-=2; n%=i;

            if(n==0)n=i;

            cout<<f[n]<<endl;

        }

        else cout<<f[n]<<endl;

    }

    return 0;

}