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HDU 1865
1sting
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3200 Accepted Submission(s): 1230
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
128
Author
z.jt
Source
2008杭电集训队选拔赛——热身赛
简单的动态规划
结果是斐波那契数列
分析:
假设长度为n的数字结果为f(n)
则f(n+1)有两种可能
1,合并后末尾依旧为1,则种类为f(n)
2,合并后末尾为2,则总类为f(n-1)
所以:f(n+1)=f(n)+f(n-1);
然后最大长度为200
直接用c++
long long 都存不下
看到java里有BigInteger类
打算来用Java玩玩
注意主类命名为Main
不能直接加 ,改为add
不能直接赋值,调用valueOf();
OK
import java.math.BigInteger;import java.util.Scanner;public class Main { public static void main(String args[]) { Scanner input = new Scanner(System.in); BigInteger ans[] = new BigInteger[201]; ans[0] = BigInteger.valueOf(1);; ans[1] = BigInteger.valueOf(1);; for(int i = 2;i<201;i++) { ans[i]=ans[i-1].add(ans[i-2]); } int T=input.nextInt(); while(T!=0) { String str = input.next(); int len = str.length(); System.out.println(ans[len]); T--; } }}
HDU 1865
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