Children’s Queue - HDU 1297 简单DP
2015-04-29 17:54
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Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11915 Accepted Submission(s): 3874
Problem Description
There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side.
The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)
Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.
Sample Input
1
2
3
Sample Output
1
2
4
题意:n个同学站队,女生不能落单,问有多少种排列方案
思路:dp[i][1]表示第i位是男生的方案数,dp[i][0]表示第i位是女生的方案数,转移方程为
dp[i][1]=dp[i-1][1]+dp[i-1][0];
dp[i][0]=dp[i-2][1]+dp[i-1][0];
Hint:注意数据范围,要用高精度,求方便直接用了Java的BigInteger
import java.util.Scanner;
import java.math.*;
public class HDU1297 {
public static void main(String[] args) {
int n;
BigInteger dp[][]=new BigInteger [1010][2];
dp[1][1]=new BigInteger("1");
dp[1][0]=new BigInteger("0");
dp[2][1]=new BigInteger("1");
dp[2][0]=new BigInteger("1");
for(int i=3;i<=1000;i++)
{
dp[i][1]=dp[i-1][1].add(dp[i-1][0]);
dp[i][0]=dp[i-2][1].add(dp[i-1][0]);
}
Scanner in=new Scanner(System.in);
while(in.hasNext())
{
n=in.nextInt();
System.out.println(dp
[1].add(dp
[0]));
}
}
}
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