HDU_1005_Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 176530    Accepted Submission(s): 43689

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).

 

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 31 2 100 0 0

 

Sample Output

25

 

本题解题思路：

通常此类型的题目有一些特点：

1、公式固定。

2、n非常大。

由此判断应该找规律求解。

然而这道题的一个比较困难的地方就在于A和B是不固定的，因此需要自己找出循环节求解。

至于循环节的求法，O__O "…，参考了下他人的做法，较简单的是：因为已知f(1)=1,f(2)=1，所以可以判断所求的f(n)和f(n-1)是否都为1，如果都为1，那么就找到了循环节。（循环节为n-2）

#include<iostream> #include<cmath> #include<cstring> using namespace std; #define maxn 100 int ans[maxn]; int a,b,n; int main(){ while(cin>>a>>b>>n){ memset(ans,0,sizeof(ans)); if(a==0&&b==0&&n==0) break; ans[1]=1; ans[2]=1; int i; for(i=3;i<maxn;i++){ ans[i]=(a*ans[i-1]+b*ans[i-2])%7; if(ans[i]==1&&ans[i-1]==1){ break; } } i=i-2; int circle=n%i; if(circle==0) cout<<ans[i]<<endl; else cout<<ans[circle]<<endl; } return 0; } </cstring></cmath></iostream>