Project Euler 78:coin Partitions

Coin Partitions

Let P (n) represent the number of different ways in which n coins can is separated into piles. For example, five coins can separated to piles in exactly seven different ways, so P (5) =7.

Ooooo
Oooo O
OOO OO
OOO o O
Oo Oo O
OO o o O
o o O-o

Find the least value of N for which P (n) are divisible by one million.

Coin Split

Note P (n) is the number of different ways to split n coins into piles. For example, five coins have 7 different ways of splitting piles, so P (5) = 7.

Ooooo
Oooo O
OOO OO
OOO o O
Oo Oo O
OO o o O
o o O-o

Find the smallest n value that enables P (n) to be divisible by 1 million.

Ideas:

How many kinds of splitting are there?

And judge whether it can be divisible by 1 million.

Reference: wiki, Partitionfunctionp

Law One:

According to this equation:

High-energy warning:

1. Here is a two-part and

2. When the first one does not meet the conditions, namely: N<k (3k-1)/2, the second one must not be established

3. The first satisfies the condition, the second may not satisfy the condition, here is said the condition is the array subscript cannot cross the bounds

4. All conditions must be calculated, only when the first one does not meet the conditions of the current cycle

5. The front ( -1) ^ (k+1), to multiply, expand the calculation, is to calculate the matching conditions of the array

Key program:

 for (k=1;k<=n;k++) {                = k* (3*k-1)/2;                 = gk1+k;                 if  Break ;                    Plist.set (N,plist.get (n)+flag*plist.get (n-gk1))                    ; if (gk2<=N) {                    plist.set (n,plist.get (n)+flag*plist.get (n-gk2));                    }                    Plist.set (N,plist.get (n)%limit);                                        Flag*=-1;            }

Here because I just saw in the above solution expression, cause I have been engaged for a long time did not get out, no culture is terrible

Law II:

See there's no problem here.

See here, directly according to the above expression solution, but here K is not from 1-n, here I understand wrong, thought to use it.

The method above does not work, the following methods are not, really wasted a lot of time

The following procedure has a process of seeking k, here is the true meaning Ah!!!

Key program:

 while (gk<=N) {                = (i%4>1)? -1:1;                Plist.set (N,plist.get (n)+flag*plist.get (n-gk));                Plist.set (N,plist.get (n)%limit);                I++                ; int  k= (i%2==0) i/2+1:-(i/2+1);                 = k* (3*k-1)/2;            }

Java Program:

 PackageLevel3;Importjava.util.ArrayList; Public classpe078{voidrun () {intLimit = 1000000;    PARTITIONS2 (limit); }    voidPartitions2 (intlimit) {ArrayList<Integer> plist =NewArraylist<integer>(); Plist.add (1); intn = 1;  while(true){            intGK1 =1; intGk2 =2; intK=1; Plist.add (0);//Initial Nth            intFlag = 1;  for(k=1;k<=n;k++) {Gk1= k* (3*k-1)/2; Gk2= gk1+K; if(gk1>n) Break; Plist.set (N,plist.get (n)+flag*plist.get (ngk1)); if(gk2<=N) {Plist.set (N,plist.get (n)+flag*plist.get (ngk2)); } plist.set (N,plist.get (n)%limit); Flag*=-1; }            if(Plist.get (n) ==0)                 Break; N++;    } System.out.println (n); }//55374//running Time=0s784ms    voidPARTITIONS1 (intlimit) {ArrayList<Integer> plist =NewArraylist<integer>(); Plist.add (1); intn = 1; intFlag;  while(true){            intGK = 1; inti = 0; Plist.add (0);  while(gk<=N) {Flag= (i%4>1)? -1:1; Plist.set (N,plist.get (n)+flag*plist.get (ngk)); Plist.set (N,plist.get (n)%limit); I++; intk= (i%2==0)? i/2+1:-(i/2+1); GK= k* (3*k-1)/2; }                        if(Plist.get (n) ==0)                 Break; N++;    } System.out.println (n); }//55374//running time=1s155ms     Public Static voidMain (string[] args) {LongT0 =System.currenttimemillis (); NewPE078 (). run (); LongT1 =System.currenttimemillis (); Longt = T1-t0; System.out.println ("Running Time=" +t/1000+ "s" +t%1000+ "MS"); }}

Fahsarm

and gives a way to ask for K.

Key program:

 while True:             = k * (3 * k-1)//2            = n- GK            If i < 0:                break            
    += ( -1) * * (k * k + 1) * p[i            ]ifelse 1 -        k p.append (PT)

Python program:

ImportTime ;defpartitions (limit): P= [1, 1, 2] n= 2 whiletrue:n+ = 1PT=0 I=0 K= 1 whileTrue:gk= k * (3 * k-1)//2I= N-GKifI <0: BreakPT+ = ( -1) * * (k * k + 1) *P[i] k=-1 * kifK > 0Else1-k P.append (PT)ifpt% Limit = =0:Print "n ="N"\ n"+"partition =", PT Break        if __name__=='__main__': T0=time.time () limit= 1000000partitions (limit) T1=time.time ()Print "running Time=", (T1-T0),"s"#n = 55374#running Time= 21.3049998283s

Description: Only the first method is my own writing, the other is seen in the Forum, self-organized

Project Euler 78:coin Partitions