题目:
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
显然,这是一个找周期的问题。
如果oj后台水的话,以7*7 = 49为周期,可能就直接过了。
但看到一位大佬写的,应该是以49*7 = 366为周期
AC代码:
# include <stdio.h> int main(void) { int a, b, n, t, t1, temp; while (~ scanf("%d %d %d", &a, &b, &n)) { if (a == 0 && b == 0 && n == 0) break; if (a % 7 == 0 && b % 7 == 0) // 如果都是7的倍数,对7取余都为0 所有f(3), f(4)等等都是0 { // 只有f(1), f(2), 有初值,都为1,输出1即可。 if (n < 3) printf("1\n"); else printf("0\n"); continue; } t = 1, t1 = 1; n = (n+333)%336 + 1; // n=3时336%336=0+1=1 执行一次 n=4执行二次,以此类推336次一个循环 while (n --) { temp = (a*t + b*t1) % 7; // 存要求的值。 t1 = t; t = temp; } printf("%d\n", t); } return 0; }
给一个链接, 里面有详细的解析:
https://blog.csdn.net/nameofcsdn/article/details/52708270