【杭电1005】Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158133    Accepted Submission(s): 38729

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

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可以发现一个规律：因为对7取模，f[n]不超过7，从中间或者开头开始形成一个周期，找到这个周期循环就可以

code:

<span style="font-size:14px;">#include<cstdio>long long f[100000000];int main(){long long n;int a,b,j;while(scanf("%d%d%lld",&a,&b,&n)!=EOF&&a!=0&&b!=0&&n!=0){f[1]=1;f[2]=1;int s=0;//记录周期 for(int i=3;i<=n;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;for(j=2;j<i;j++){if(f[i]==f[j]&&f[i-1]==f[j-1])//确定前两个找到一个周期{s=i-j;break;} }if(s>0)break;}if(s>0)f[n]=f[(n-j)%s+j];//周期循环并加上前面不参加循环的数printf("%lld\n",f[n]);}return 0; } </span>

另一种code：

#include<iostream>  using namespace std;  int f(int A, int B,int n)  {      if(n==1 || n==2)          return 1;      else          return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;  }  int main()  {      int a,b,n;      while(cin>>a>>b>>n,a||b||n)      {          cout<<f(a,b,n%49)<<endl;      }      return 1;  }