HDOJ 1005 Number Sequence

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本文介绍了一种通过递推公式计算特定数列中第n项值的算法实现。该算法利用模运算解决大规模数值计算问题，并通过寻找周期性规律来优化计算效率。

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

传送门

#include<iostream>
using namespace std;

int main()
{
    int a[100],A,B,I,n;
    a[1]=1;
    a[2]=1;
    while(cin>>A>>B>>n)
    {
        if(A==0&&B==0&&n==0)    break;
        int cnt=0;
        for(i=3;i<=99;i++)
        {
            a[i]=(A*a[i-1]+B*a[i-2])%7;
            if(a[i]==1&&a[i-1]==1)  break;
            if(a[i]==0&&a[i-1]==0) {cnt=1;break;}
        }
        if(cnt&&n>i-2){cout<<"0"<<endl;continue;}
        i-=2;
        n%=i;
        if(n==0)n=i;
        cout<<a[n]<<endl;
    }
}