# HDOJ 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 3
1 2 10
0 0 0

Sample Output
2
5

```#include<iostream>
#include<stdio.h>
using namespace std;
int f[100000005];
int main()
{
int a,b,n,i,j;

f[1]=1;f[2]=1;
while(scanf("%d%d%d",&a,&b,&n))
{
int s=0;//记录周期
if(a==0&&b==0&&n==0) break;
for(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-1]==f[j-1]&&f[i]==f[j])//此题可以这样做的原因就是 2个确定后就可以决定后面的
{
s=i-j;
//cout<<j<<" "<<s<<" >>"<<i<<endl;
break;
}
if(s>0) break;
}
if(s>0){

f[n]=f[(n-j)%s+j];
//cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl;
}
cout<<f[n]<<endl;

}
return 0;
}
```
```import java.util.Scanner;

public class Main {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int[] a = new int[54];
int A  = sc.nextInt();
int B  = sc.nextInt();
int n  = sc.nextInt();
if(A==0&&B==0&n==0){
return ;
}
a[1]=1;
a[2]=1;
int k=0;
int i;
for(i=3;i<54;i++){
a[i] = (A*a[i-1]+B*a[i-2])%7;
if(i>5&&a[i]==a[4]&&a[i-1]==a[3]){
k=i-4;
break;
}
}
//System.out.println(k);
if(n>2){
System.out.println(a[(n-3)%k+3]);
}else{
System.out.println("1");
}

}
}

}

```
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