Number Sequence

[align=left]Problem Description[/align]
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).

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, print the value of f(n) on a single line.

[align=left]Sample Input[/align]

1 1 3
1 2 10
0 0 0

[align=left]Sample Output[/align]
[align=left] [/align]
2

5

参考别人的代码写的，希望有一天可以不参考别人的独立解决每道题。

#include<stdio.h>
//long f[1000000000];因为周期性的存在，所以不需要把数组开得这么大！
int f[210];
int main()
{
int a,b;
long n;
while(scanf("%d%d%ld",&a,&b,&n),a||b||n)
{
if(n==1||n==2)
{
printf("1\n");
continue;
}
int i,flag=1;
f[1]=1;
f[2]=1;
for(i=3;i<=200;i++)//打表，找到一个周期就结束！
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i]==1&&f[i-1]==1)//出现周期
break;
if(f[i]==0&&f[i-1]==0)//这里有一个陷阱，当 a=7,b=7 时，后面都是零
{
flag=0;
break;
}
}
if(flag==0)
{
printf("0\n");
continue;
}
i=i-2;//i是周期
if(n<i)
printf("%d\n",f
);
else
{
n=n%i;
if(n==0)
printf("%d\n",f[i]);
else
printf("%d\n",f
);
}
}
return 0;
}