# HDU 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

#include <iostream>
using namespace std;
int f[1000];//定义在外面范围大一些;
int main()
{
int a,b,n;
while(cin>>a>>b>>n)
{
if(a==0&&b==0&&n==0)
break;
f[1]=1;
f[2]=1;
int flag;
for(int i=3;i<=1000;i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i]==f[i-1])//对于每一个数组,前两个元素大小都是1,都是相等的，所以可以作为判断是否循环的条件;
{
flag=i-2;//如果满足这个条件的话,i-1是找到的下一个循环节的起始位置,那么i-2就是上一个循环的结束;
}
}
cout<<f[n%flag]<<endl;//flag就是循环周期;
}
return 0;
}

```

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