杭电1005-Number Sequence

问题描述

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).

输入说明

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.

输出说明

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

输入样例

1 1 3
1 2 10
0 0 0

输出样例

2
5









这道题之前用递归做的，结果runtimeError了，估计递归非常慢，想想会超出堆栈.后来百度，原来是个循环节的概念，因为有mod 7，所以f（n）取值是0-6,7个取值，而f(n)又由上头两个决定，因此有7*7=49种答案，因此在50以内必然出现循环，所以我们用数组模拟前49组数组，后面的数据只要mod （模除）循环节就可以了，对应的的数组里头取值，下面附上代码




------------------------------------下面附上AC代码-----------------------

#include <iostream>
using namespace std;
int main()
{
        int A,B;
        int N;
        int f[50];
        while(cin>>A>>B>>N)
        {
                if(A==0&&B==0&&N==0)break;
                int i;
                f[1]=f[2]=1;
                for(i=3;i!=50;++i)
                {
                        f[i]=(A*f[i-1]+B*f[i-2])%7;
                        if(f[i]==1&&f[i-1]==1)break;
                }
                N=N%(i-2);//理解i-2，求的循环节N
                if(N==0)cout<<f[(i-2)]<<endl;
                else cout<<f[N]<<endl;
        }
        
        return 0;
}