UVA10334 Ray Through Glasses【大数】

Suppose we put two panes of glass back-to-back. How many ways an are there for light rays to pass through or be reflected after changing direction n times? Following figure shows the situations when the value of n is 0,1 and 2.

Input

It is a set of lines with an integer n where 0 ≤ n ≤ 1000 in each of them.

Output

For every one of these integers a line containing an as described above.

Sample Input

0

1

2

Sample Output

1

2

3

题链接：UVA10334 Ray Through Glasses

问题简述：（略）

问题分析：

  这个问题其实就是fibonacci数列，只是起始项的值不同。递推式如下：

  f0=0

  f1=2

  ......

  fn=fn-2 + fn-1(n>=2)

  大数计算是用模拟法来实现的，参见程序。

程序说明：

  用Java程序来完成大数计算是最为方便的。

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

AC的C++语言程序（模拟法）如下：

/* UVA10334 Ray Through Glasses */

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ULL;

const int N = 1000;
const int N2 = 1000 / 17;
const ULL BASE = 1e17;
ULL fib[N + 1][N2 + 1];

void setfib()
{
    fib[0][0] = 1;
    fib[1][0] = 2;
    int len = 1;
    for(int i = 2; i <= N; i++) {
        int carry = 0;
        for(int j = 0; j < len; j++) {
            fib[i][j] = fib[i - 2][j] + fib[i - 1][j] + carry;
            carry = fib[i][j] / BASE;
            fib[i][j] %= BASE;
        }
        if(carry)
            fib[i][len++] = carry;
    }
}

int main()
{
    memset(fib,sizeof(fib));

    setfib();

    int n;
    while(~scanf("%d",&n)) {
        int k = N2;
        while(fib[n][k] == 0)
            k--;
        printf("%lld",fib[n][k]);
        for(int i = k - 1; i >= 0; i--)
            printf("%017lld",fib[n][i]);
        printf("\n");
    }

    return 0;
}

AC的Java语言程序如下：

/* UVA10334 Ray Through Glasses */

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

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(system.in);
        BigInteger[] fib = new BigInteger[1001];
        fib[0] = new BigInteger("1");
        fib[1] = new BigInteger("2");
        for(int i = 2; i <= 1000; i ++)
            fib[i] = fib[i - 1].add(fib[i - 2]);
        while(cin.hasNext())
        {
            int N = cin.nextInt();
            System.out.println(fib[N]);
        }
    }
}