ACM--递推加大数--HDOJ 1865--1string--水

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HDOJ题目地址：传送门

1sting

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5135 Accepted Submission(s): 1926

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

311111111

Sample Output

分析(转)：

简单递推+大数。
f[n]=f[n-1]+f[n-2]。
由于数字只有'1'和'2'这两种，那么当让第n
位为1的时候，f[n]加上f[n-1]；当让最后两位
合并成2的时候，f[n]加上f[n-2]（因为要保证
倒数第二位一定是1）。

算出来的f就是一个斐波那契数列。

import java.math.BigDecimal;import java.math.BigInteger;import java.util.Scanner;public class Main {   public static void main(String[] args) {    //f(n)=f(n-1)+f(n-2)      Scanner cin= new Scanner(System.in);      int n=cin.nextInt();      BigDecimal []big=new BigDecimal[300];      big[1]=new BigDecimal(1);      big[2]=new BigDecimal(2);      for(int i=3;i<300;i++){          big[i]=big[i-1].add(big[i-2]);      }      while(n-->0){          String s=cin.next();          int len=s.length();          System.out.println(big[len].toPlainString());      }   }}

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