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).
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> int main() { int n; int A,B; int a[10000]= {0,1,1}; while(scanf("%d%d%d",&A,&B,&n)&&(A||B||n)) { int i; for(i=3; i<10000; i++) { a[i]=(A*a[i-1]+B*a[i-2])%7; if(a[i-1]==1&&a[i]==1) break; } n=n%(i-2); if(n==0) n=i-2; printf("%d ",a[n]); } return 0; }
因在计算过程中涉及求余,可利用周期性及时跳出循环