HDOJ/HDU 1865 1sting(斐波拉契+大数~)

Problem Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.

Output

The output contain n lines, each line output the number of result you can get .

Sample Input

3

1

11

11111

Sample Output

1

2

8

题意：

若干个1，可以选择相邻两个合并成2。问有多少种可能的结果。

分析：递推加大数~

递推公式为db[i] = db[i-1] + db[i-2]，斐波那契数列。

怎么推导出来的呢~~~我能说我是看出来的麽~

设有n个1，可以构成f(n)种。则加一个1的时候，前面n种仍然成立 f(n+1)=f(n)+*;

第n+1个1和第n个1相加构成2，前面n-1个1可以组合的个数。 f(n+1)=f(n)+f(n-1);

大数~用java很好过的~c的话，只能用数组模拟了。

import java.math.BigInteger;
import java.util.Scanner;
public class Main{
    static BigInteger db[] = new BigInteger[201];
    public static void main(String[] args) {
        dabiao();
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            String str =sc.next();
            int n =str.length();
            System.out.println(db[n]);
        }
    }
    private static void dabiao() {
        db[1]=new BigInteger("1");
        db[2]=new BigInteger("2");
        for(int i=3;i<db.length;i++){
            db[i]=db[i-1].add(db[i-2]);
        }
    }
}