（递推大整数） Children’s Queuehdu1297

mac2022-06-30 107

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 16006 Accepted Submission(s): 5337

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

Sample Output

对于各种排列而言，用f(n)表示其长度为n时的可能数，有以下的情形：

　　长度为1时，只有1种可能，即“M”（女生不能单独站立，下同）；

　　长度为2时，有2种可能，即“FF”和“MM”;

　　长度为3时，有4种可能，即“FFF”、“FFM”、“MFF”和“MMM”;

　　长度为4时，有7种可能，即“FFFF”、“FFFM”、“FFMM”、“MFFM”、“MFFF”、“MMFF”、“MMMM”;

　　长度为n>4时，对于之前长度n-1的排列，加一个M是可行的（最好是男孩的情形）；对于之前长度n-2的排列，加加上FF是可行的（加MM有可能造成排列重复，这是最后是女孩的情形，并且前n-2个人是可行的排列）；另外一种情况是，前n-2个人是不可行的排列，即最后的两个人是“MF”，再加上两个“FF”就变成可行的排列，这种情况也就是在n-4人可行的排列基础上加上“MFFF”。

C++：

#include<iostream> #include<cstring> #include<cstdio> using namespace std; int s[1001][200]; int main() { s[1][0]=1; s[2][0]=2; s[3][0]=4; s[4][0]=7; int i,j,c,ans=0; for(i=5;i<=1000;i++) { for(j=0,c=0;j<200;j++) { ans=s[i-1][j]+s[i-2][j]+s[i-4][j]+c; c=ans/100000000; s[i][j]=ans%100000000; } } int n; while(cin>>n) { j=199; while(!s[n][j]) j--; cout<<s[n][j]; for(i=j-1;i>=0;i--) printf("d",s[n][i]); cout<<endl; } return 0; }

用JAVA

import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner (System.in); BigInteger a[]=new BigInteger[1001]; int n; while(in.hasNextInt()) { n=in.nextInt(); a[1]=BigInteger.valueOf(1); a[2]=BigInteger.valueOf(2); a[3]=BigInteger.valueOf(4); a[4]=BigInteger.valueOf(7); for(int i=5;i<=n;i++) { a[i]=BigInteger.ZERO; a[i]=a[i].add(a[i-1]); a[i]=a[i].add(a[i-2]); a[i]=a[i].add(a[i-4]); } System.out.println(a[n]); } } }

转载于:https://www.cnblogs.com/Weixu-Liu/p/9164693.html

（递推 大整数） Children’s Queuehdu1297

（递推大整数） Children’s Queuehdu1297