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
序列是循环的,以49为周期,具体原因不明
#include<iostream> using namespace std; void solve() { int A, B, n; while(cin>>A>>B>>n && (A || B || n)) { int a[2] = {1, 1}; for(int i = 0; i < (n %49 - 1) / 2; i++) { a[0] = (A * a[1] + B * a[0]) % 7; a[1] = (A * a[0] + B * a[1]) % 7; } if(n % 2) { cout<<a[0]<<endl; } else { cout<<a[1]<<endl; } } } int main() { solve(); return 0; }
作者:xueda120
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