题目：
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).
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.
# include <iostream> # include <string> using namespace std; /*因为用到了迭代的方法，一定要控制的迭代的次数，否则会出现栈溢出的情况。 分析问题可以推断，由于f(n)是由前两个数字组合产生，那么只要有两个数字组合相同的情况发生就一定一会产生循环！ 两个数字的组合的最大可能值为7x7=49，因此只要在调用迭代方法中限制n的在0~48就可以了。 自认为本人这个程序比较容易接受，是迭代的方法。*/ int function(int a, int b, int n){ if (n==1) return 1; else if(n==2) return 1; else return (a*function(a, b, n-1)+b*function(a, b, n-2))%7; } int main(){ int a, b, n; while(cin >> a >> b >> n){ if(a == 0 && b == 0 && n == 0) break; else cout << function(a, b, n%49) << endl; } system("pause"); return 0; }