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
题解：打表找规律
感受：将for循环里面的1000换成maxn，会wrong，想不通。将for换成while也会wrong，懵逼!!!!!!
1 #pragma warning(disable:4996) 2 #include<map> 3 #include<string> 4 #include<cstdio> 5 #include<bitset> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 typedef long long ll; 11 12 const int maxn = 1005; 13 14 int a, b, n; 15 int F[maxn]; 16 17 int main() 18 { 19 F[1] = F[2] = 1; 20 while (scanf("%d%d%d", &a, &b, &n) != EOF) { 21 if (!a && !b && !n) break; 22 int k; 23 for (k = 3; k < 1000; k++) { 24 F[k] = (a*F[k - 1] + b * F[k - 2]) % 7; 25 if (F[k] == 1 && F[k - 1] == 1) break; 26 } 27 n = n % (k - 2); 28 F[0] = F[k - 2]; 29 printf("%d\n", F[n]); 30 } 31 return 0; 32 }
