ACM一句代码求解释题目ProblemDescriptionAnumbersequenceisdefinedasfollows:f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7.GivenA,B,andn,youaretocalculatethevalueoff(n).InputTheinputconsistsofmultipletes

Question

ACM一句代码求解释
题目Problem 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

Answer 1

代码格式没有细究,思路方向是对的.如果f(n) = f(m) 且 f(n+1) = f(m+1) 则可断定 f(n+p) = f(m+p) (p>=0).f(n)取值情形不会超过7种,f(n)和f(n-1)的取值组合情形不超过49种,即循环一定存在.设k为循环周期,f(n)=f(n+k)...