【规律题】

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/*Number SequenceTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 102269    Accepted Submission(s): 24704Problem DescriptionA 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).InputThe 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. OutputFor each test case, print the value of f(n) on a single line.Sample Input1 1 31 2 100 0 0Sample Output25 AuthorCHEN, Shunbao SourceZJCPC2004 */#include<stdio.h>int f(int a, int b, int n){if(n == 1 || n == 2) return 1; else return ( ( (a % 7) * (f(a, b, n-1) %7) ) + ( (b%7) * (f(a, b, n-2)  %7) ) )% 7;}//递归次数过多时， 堆栈会溢出int main(){int a, b, n;while(scanf("%d%d%d", &a, &b, &n) != EOF && (a || b || n))printf("%d\n", f(a,b, n % 49));//49规律控制递归次数return 0;}

题意：给出一个函数f（n），输入n输出f（n）。

思路：这道题如果单纯的用函数递归来做会爆栈，需要找出这个函数序列的循环节，49一个循环，然后mod求解即可。

f(n)的值一定是0-6之间的一个数，共7种情况，f(n-1)和f(n-2)同样如此。所以，在A和B固定的前提下，A * f(n - 1) + B * f(n - 2)的组合有7×7种情况。也就是说，最多有49种组合方式。因此，超过49以后，必定存在循环。

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