[Recurrence of large numbers] HDU 1297 -- children's queue

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The recursive problem is easy to implement, but it is really troublesome to get the recursive formula, just like DP.

Analysis (some ppt files from HDU ):

Set: F (n) indicates the valid queue of n persons, then:
According to the gender analysis of the last person, he is either male or female, so he can be divided into two categories:
1. If the last person in the valid queue of N people is male, there is no restriction on the queue of n-1 people. He only needs to stand at the end, in this case, F (n-1) is used );

2. If the last person in the valid queue of n persons is a female, the n-1 person in the queue must be a girl. That is to say, the last two persons must be both girls, this can be divided into two situations:
(1), if the queue of the former N-2 individual is legal queue, then obviously followed by two girls, it must be legal, this situation has F (n-2 );

(2) the difficulty is that, even if the previous N-2 individual is not legal queue, plus two girls may also be legal, of course, this length for the N-2 of illegal queue, the invalid part must be the tail, that is to say, the length here is the form of the illegal string of N-2 must be "F (n-4) + male + female ", this case has a total of F (n-4 ).

So the recursive formula is F (n) = f (n-1) + f (n-2) + f (n-4 ).

In addition, it should be noted that the recursive data range is 1000, and Int or even long can no longer carry the maximum boundary. Therefore, I used the large number class posted two days ago.

# Include <iostream> # include <string> # include <iomanip> # include <algorithm> using namespace STD; # define maxn 9999 # define maxsize 10 # define dlen 4 class bignum {PRIVATE: int A [500]; // The number of digits that can be controlled for a large number int Len; // the length of a large number is public: bignum () {Len = 1; memset (A, 0, sizeof (a);} // constructor bignum & operator = (const bignum &); // reload the value assignment operator. bignum operator + (const bignum &) const; // reload the addition operator. Void print () is an addition operation between two large numbers (); // output large number}; Bignum & bignum: Operator = (const bignum & N) // overload the value assignment operator. assign values between large numbers {int I; Len = n. len; memset (A, 0, sizeof (a); for (I = 0; I <Len; I ++) A [I] = n. A [I]; return * This;} bignum: Operator + (const bignum & T) const // The addition operation between two large numbers {bignum T (* This ); int I, big; // The number of digits big = T. len> Len? T. len: Len; for (I = 0; I <big; I ++) {T. A [I] + = T. A [I]; If (T. A [I]> maxn) {T. A [I + 1] ++; T. A [I]-= maxn + 1 ;}} if (T. A [Big]! = 0) T. len = big + 1; elset. len = big; return t;} void bignum: Print () // output large number {int I; cout <A [Len-1]; for (I = len-2; I> = 0; I --) {cout. width (dlen); cout. fill ('0'); cout <A [I];} cout <Endl;} bignum que [1202]; // The above is not the main function .. Int main () {que [0] = 1; que [1] = 1; que [2] = 2; que [3] = 4; for (INT I = 4; I <= 1000; I ++) {que [I] = que [I-1] + que [I-2] + que [I-4];} int tar; while (CIN> TAR) {que [tar]. print ();} return 0 ;}