这个是我当年AC的C++版本:
#include<iostream>
using namespace std;
int main()
{
int A,B;
unsigned int n;
cin>>A>>B>>n;
while(A&&B&&n)
{
int *p=new int[50]; //存在循环,在这里判断,来减少运行时间
if(n>50)
n=(n-2)%48+2;
*p=*(p+1)=1;
for(int i=2;i<n;i++)
*(p+i)=(A**(p+i-1)+B**(p+i-2))%7; //公式
cout<<*(p+n-1)<<endl;
cin>>A>>B>>n;
}
return 0;
}
#include<iostream>
using namespace std;
int main()
{
int A,B;
unsigned int n;
cin>>A>>B>>n;
while(A&&B&&n)
{
int *p=new int[50]; //存在循环,在这里判断,来减少运行时间
if(n>50)
n=(n-2)%48+2;
*p=*(p+1)=1;
for(int i=2;i<n;i++)
*(p+i)=(A**(p+i-1)+B**(p+i-2))%7; //公式
cout<<*(p+n-1)<<endl;
cin>>A>>B>>n;
}
return 0;
}
import java.util.Scanner;public class NumberSequence2 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a, b;
int n;
while (in.hasNext()) {
a = in.nextInt();
b = in.nextInt();
n = in.nextInt();
// 1 <= A, B <= 1000, 1 <= n <= 100,000,000
if (a < 1 & a > 1000 & b < 1 & b > 1000 & b < 1 & a > 100000000)
System.exit(0);
if (a == 0 & b == 0 & n == 0)
System.exit(0);
int f[] = new int[50];
for (int i = 1; i < 50; i++) {
if (i == 1 || i == 2) {
f[i] = 1;
} else {
f[i] = (a * f[i - 1] + b * f[i - 2]) % 7;
}
}
System.out.println(f[n % 49]);
}
in.close();
}}
另附本人关于此题的博客一篇,欢饮讨论:
http://blog.csdn.net/jinyongqing/article/details/21537175