Number Sequence

http://acm.hdu.edu.cn/showproblem.php?pid=1005

找循环节。。

#include <stdio.h>
#include <string.h>
const int N=52;
int f[N];
int main()
{
    int a,b,n;
    while(~scanf("%d%d%d",&a,&b,&n))
    {
        int s,e;
        if (a==0&&b==0&&n==0)
            break;
        f[0] = 1;
        f[1] = 1;
        f[2] = 1;
        for (int i = 3; i <= n; i++)
        {
            f[i] = (a*f[i-1]+b*f[i-2])%7;
            for (int j = 2; j < i; j++)
            {
                if (f[i]==f[j]&&f[i-1]==f[j-1])
                {
                    s = j;
                    e = i;
                    goto RL;
                }
            }
        }
        printf("%d
",f[n]);
        continue;
RL:
        printf("%d
",f[s+(n-e)%(e-s)]);
    }
    return 0;
}

用java写了一下，果断的超时了。。

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner cin = new Scanner(System.in);
        BigInteger M = BigInteger.valueOf(7);
        while (cin.hasNext()) {
            BigInteger a = cin.nextBigInteger();
            BigInteger b = cin.nextBigInteger();
            BigInteger n = cin.nextBigInteger();
            int s1 = a.compareTo(BigInteger.ZERO);
            int s2 = b.compareTo(BigInteger.ZERO);
            int s3 = n.compareTo(BigInteger.ZERO);
            if (s1 == 0 && s2 == 0 && s3 == 0)
                break;
            BigInteger f1 = BigInteger.ONE;
            BigInteger f2 = BigInteger.ONE;
            BigInteger cnt = BigInteger.valueOf(2);
            BigInteger f;
            while (true) {
                f = a.multiply(f2).add(b.multiply(f1)).mod(M);
                f1 = f2;
                f2 = f;
                cnt = cnt.add(BigInteger.ONE);
                s1 = n.compareTo(cnt);
                if (s1 == 0)
                    break;
            }
            System.out.println(f);

        }
    }

}