HDU1297 Girl can't Walk Alone (DP)

Children ' s Queue

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13333 Accepted Submission (s): 4364

Problem Descriptionthere is many students in PHT School. One day, the headmaster whose name was Pigheader wanted all students stand in a line. He prescribed that girl can is in a single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n was 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 the queue satisfied the headmaster ' s needs is 7. Can do a program to find the total number of the queue with n children?

Inputthere is multiple cases in this problem and ended by the EOF. In each case, there are only one integer n means the number of children (1<=n<=1000)

Outputfor Each test case, there are only one integer means the number of the queue satisfied the headmaster ' s needs.

Sample Input123

Sample Output124

Authorsmallbeer (CML)

SOURCE Hangzhou Electric ACM Training Team Training Tournament (VIII)

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A[I-1]: Legal + male
A[I-2}: Legal + female
A[I-4}: Legal + unisex (i.e. not legal) + female

Here are three different situations that don't cross each other!

Explain:
Set: F (n) represents the legal queue for n individuals:

According to the last person's gender analysis, he is either male or female,
So there are two main types of discussions:

1. If the last person in the legal queue of n person is a male,
The front n-1 personal queue as long as it is a legitimate queue,
The last guy just needs to stand at the end, so this is a total of f (n-1);

2. If the last person in the legal queue of n person is a female,
The n-1 person in the queue must also be a schoolgirl, which means
Limit the last two people must be girls, this can be divided into two kinds of situations:

2.1. If the front n-2 person of the queue is a legitimate queue,
It is obvious that two more girls will be added to the rear, and it must be legal, this situation has f (n-2);
2.2, however, even if the front n-2 individual is not a legal queue, plus two girls may also be legal,
Of course, this is the n-2 of the length of the illegal queue, the illegal place must be the tail,
That is, the length of the n-2 is the illegal string of the form must be f (n-4) + male + female,
There is a total of f (n-4) in this case.
Source

HDU1297

Ideas are as follows:

A queue of length n can be seen as an additional 1 children in a n-1 queue, the child may only be:

A. Boys, the addition of 1 boys to any legal queue of n-1 is necessarily legal and the number of cases is f[n-1];

B. Girls, the addition of 1 girls in front of the queue after the end of the n-1 to the girl is also legal, we can convert the 2-bit girls into the n-2 queue;

One case is the addition of 2 girls in the legal queue of n-2, the number of cases is f[n-2];

But we notice the difficulty of the subject, perhaps the former n-2 bit with the girl as the end of the illegal queue (that is, the mere 1 girls end),
It is also possible to append 2 girls to the legal queue, while this n-2 illegal queue is bound by the structure of n-4 legal queue + 1 boys +1 girls,
That is, the number of cases is f[n-4].

If you feel the subject is more difficult, you can first review the article: [Acm_hdu_2045]lele's RPG puzzle, the idea is similar to the subject, but relatively simple.
A recursive formula is obtained as follows:
A[I]=A[I-1]+A[I-2]+A[I-4];

#include <iostream>#include<stdlib.h>#include<stdio.h>using namespacestd;Long Longa[10005][ the]={0};intMain () {intI,j,n; a[1][0]=1; a[2][0]=2; a[3][0]=4; a[4][0]=7; a[5][0]= A;  for(i=6; i<= +; i++)    {         for(j=0; j<= the; j + +) {A[i][j]+=a[i-1][j]+a[i-2][j]+a[i-4][j]; A[i][j+1]+=a[i][j]/100000000; A[I][J]%=100000000; }    }     while(cin>>N) { for(j= the; j>=0; j--)        if(a[n][j]!=0)         Break; cout<<A[n][j];  for(j=j-1; j>=0; j--) printf ("%08d", A[n][j]);//It's not right to cout a number of times.cout<<Endl; }    return 0;}

HDU1297 Girl can't Walk Alone (DP)