题目链接

1sting HDU - 1865

题目

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

给定一个只由“1”组成的字符串s，相邻的两个“1”可以合并成“2”。比如“1111”可以组成字符串”1111“，”121“，”112“，”211“，”22“。这5个不重复的字符串。
现要求s可以组成的字符串的个数。

分析

设F(i)表示字符串长度为i的字符串可以组成的字符串的个数。
根据题意，可以得知：
F(1)=1;
F(2)=2;
F(3)=3;
F(4)=5;
F(5)=8;
F(6)=13;
观察规律，可以得到：
F(i)=F(i-1)+F(i-2),i>2;
然后用JAVA处理大数问题即可。

AC代码

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

public class Main{
    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        int n;
        n=cin.nextInt();
        for(int i=1;i<=n;i++) {
            int x;
            String s;
            s=cin.next();
            x=s.length();
            BigInteger a,b,c;
            a=BigInteger.ONE;
            b=a.add(a);
            c=a;
            if(x==2) c=b;
            for(int j=3;j<=x;j++) {
                c=a.add(b);
                a=b;
                b=c;
            }
            System.out.println(c);
        }
    }
}