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
(a+b) mod c == (a mod c + b mod c) mod c
抽屉原理。
令[f(n-1), f(n-2)] 为一组,所以一共有 7*7 种不同的组合情况,到第 50 个时,一定会开始出现重复的组合,因为斐波那契数列 f(n) = f(n-1)+f(n-2),所以开始循环重复,所以 7*7 为最大周期。
#include <iostream> using namespace std; int main(void) { int a, b, n; int s[50]; while(cin >> a >> b >> n && (a || b || n)) { s[0] = 1; s[1] = 1; for(int i = 2; i < 49; i++) { s[i] = (a*s[i-1]+b*s[i-2]) % 7; } cout << s[n%49-1] << endl; } return 0; }