HDOJ 1005 Number Sequence

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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
发现很多同学都是以1,1为重复头，按照最多循环次数48来做的
我也参考了一些答案，发现：
1，不能以1,1 作为重复头；
2，自己先找周期。

#include<iostream>#include<stdio.h>using namespace std;int f[100000005];int main(){    int a,b,n,i,j;    f[1]=1;f[2]=1;    while(scanf("%d%d%d",&a,&b,&n))    {        int s=0;//记录周期        if(a==0&&b==0&&n==0) break;        for(i=3;i<=n;i++)        {            f[i]=(a*f[i-1]+b*f[i-2])%7;            for(j=2;j<i;j++)            if(f[i-1]==f[j-1]&&f[i]==f[j])            //此题可以这样做的原因就是 2个确定后就可以决定后面的            {                s=i-j;                //cout<<j<<" "<<s<<" >>"<<i<<endl;                break;            }            if(s>0) break;        }        if(s>0){                 f[n]=f[(n-j)%s+j];     //cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl;               }        cout<<f[n]<<endl;    }    return 0;}

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