(递推 大整数) Children’s Queue hdu1297
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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
1
2
3
Sample Output
1
2
4
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("%08d",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]); } } }
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