ACM杭电1005 Number Sequence

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

刚开始读这道题觉得很简单，就是一个公式而已，然后带进去翻译成代码即可，结果交了一次WA，然后找问题，才发现这个公式是个周期函数，于是自己打表，从1 1到10 10虽然没有全部打表，但也打了大部分，后面发现 7 7全0后面的8 8就与1 1重合了，于是感觉这是个周期函数，每套数据48个，然后就用nmod49让n能取到48。代码如下：

#include<iostream>#include<math.h>using namespace std;const int MAX_N=100000005;int f[MAX_N];int main(){int A,B,n;while (cin>>A>>B>>n){int T=0;int f1=0;f[0]=0;f[1]=1;f[2]=1;if(A==0&&B==0&&n==0)break;A=A%7;B=B%7;n=n%49;for(int i=3;i<=n;++i){f[i]=(A*f[i-1]+B*f[i-2])%7;}if(A==7&&B==7)cout<<"0"<<endl;if(A==0&&B==0)cout<<"0"<<endl;else   cout<<f[n]<<endl;}return 0;}