HDOJ 1005 Number Sequence

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
=f[(n-j)%s+j];
//cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl;
}
cout<<f
<<endl;

}
return 0;
}