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
1 2 10
0 0 0
Sample Output
2
5
5
#include<iostream>
#define N 10000
using namespace std;
int main()
{
int A,B,n;
cin>>A>>B>>n;
n=n%;
while(!(A==&&B==&&n==))
{
long long int a[N];
a[]=;
a[]=;
if(n==||n==)
cout<<<<endl;
else
{
for(int i=;i<=n;i++)
{
a[i]=(A*a[i-]+B*a[i-])%;
}
cout<<a[n]<<endl;
}
cin>>A>>B>>n;
n=n%;
}
return ;
}
由于f(n)是由前两个数字组合产生，那么只要有两个数字组合相同的情况发生就一定一会产生循环！
两个数字的组合的最大可能值为7x7=49，因此只要在调用迭代方法中限制n的在0~48就可以了