Project Euler 88:product-sum numbers product and sum

Product-sum numbers

A natural number, N, that can is written as the sum and product of a given set of at least the natural numbers, {a1, A2, ... , AK} is called a product-sum number:n = a1 + A2 + ... + ak = A1xa2x...xak.

For example, 6 = 1 + 2 + 3 = 1x2x3.

For a given set of size, K, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, K = 2, 3, 4, 5, and 6 is as follows.

K=2:4 = 2x2 = 2 + 2
K=3:6 = 1x2x3 = 1 + 2 + 3
K=4:8 = 1x1x2x4 = 1 + 1 + 2 + 4
K=5:8 = 1x1x2x2x2 = 1 + 1 + 2 + 2 + 2
K=6:12 = 1x1x1x1x2x6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; Note that 8 is a counted once in the sum.

In fact, the AS-the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 61, +, +}, the sum is.

What's the sum of the minimal product-sum numbers for 2≤k≤12000?

Accumulate and Sum

If the natural number n can be simultaneously expressed as a group of at least two natural number {A1, A2, ..., AK} the product sum and sum, also namely n = A1 + a2 + ... + ak = A1xa2x...xak, then n is called the product and the accumulation.

For example, 6 is an accumulation and sum, because 6 = 1 + 2 + 3 = 1x2x3.

Given the size of the set K, we call the minimum n value of the above property to be the minimum product sum. When k = 2, 3, 4, 5, 6 o'clock, the minimum product sum is as follows:

K=2:4 = 2x2 = 2 + 2
K=3:6 = 1x2x3 = 1 + 2 + 3
K=4:8 = 1x1x2x4 = 1 + 1 + 2 + 4
K=5:8 = 1x1x2x2x2 = 1 + 1 + 2 + 2 + 2
K=6:12 = 1x1x1x1x2x6 = 1 + 1 + 1 + 1 + 2 + 6

Therefore, for 2≤k≤6, the sum of all the minimum product counts is 4+6+8+12 = 30; Note 8 is counted only once.

For 2≤k≤12, the set of all minimum product and sum is known to be {4, 6, 8, 12, 15, 16}, these numbers are 61.

For 2≤k≤12000, what is the sum of all the minimum product numbers?

Solving

Product of the number of k and = = K

For the smallest number in the K-time.

The topic requires 2≤k≤12000, the time of the smallest plot no and

Procedures in the reference, detailed program notes

Java

 PackageLevel3;ImportJava.util.Set;ImportJava.util.TreeSet; Public classpe088{Static voidrun () {intKmin = 2; intKmax = 12000; intsum = 0; Set<Integer> set =NewTreeset<integer>();  for(intk=kmin;k<=kmax;k++){            intMinn =Getmin (k); if(SET.ADD (Minn)) sum+=Minn;    } System.out.println (sum); }    //Find k for the smallest n    Static intGetmin (intk) {         for(intn=k+1;; n++){            if(check (n,n,k))returnN; }    }    //A number of numbers that are split into k and or K numbers.//prod Multiply//Sum and//at first, these two numbers are equal, prod or sum split into K-parts.//here is the recursive method, when prod2 = Prod1 * A; sum2 = Sum1-a//The next round of check (PROD2,SUM2,K-1) can be detected//here with recursion is also because possible out of 8 = 2*2*2*1*1 = 2+2+2+1+1 form, in which there are several identical cases in the multiplicative child//End Condition: When the multiplier is 1, sum = = k K 1 sum is sums//k=1 the end of the time. return prod = = Sum//next recursion is required: d<= prod k-1<= sum-d The following procedure is obviously    Static BooleanCheckintProdintSumintk) {        if(Sum <k)return false; if(prod = = 1)returnsum==K; if(k==1)returnProd = =sum;  for(intD =2;d<= prod && sum-d>=k-1;d++){            if(prod%d==0){                if(Check (prod/d,sum-d,k-1))                     return true; }        }        return false; }//7587457//running Time=1s577ms     Public Static voidMain (string[] args) {LongT0 =System.currenttimemillis ();        Run (); LongT1 =System.currenttimemillis (); Longt = T1-t0; System.out.println ("Running Time=" +t/1000+ "s" +t%1000+ "MS"); }}

Reference links

N[k] means minimal product-sum numbers for size=k

The upper bound of N[k] is 2*k, because 2*k can always be decomposed into 2*k, then 2*k=k+2+ (1) * (k-2)

Obviously the nether of N[k] is K

For a number of num-factorization after decomposition, the number of factors is the sump of the factors of product

The number of 1 that needs to be added is num-sump, so size k=num-sump+product

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That's a good understanding.

In the case of factoring

N[K] is a number of numbers broken down into the number of K and K number of the product of the minimum value

I link above the program's understanding is through the factorization, continuously reduces the value of n[k], the final value is the smallest, but the recursion after the program is not understood ...

#CODING=GBKImport time as timedefrun2 (): KMax= 12000N= [2*kmax forIinchrange (KMax)]defGETPSN (Num,sump,product,start): K= Num-sump +ProductifK <KMax:ifnum<N[k]: n[k]=Num forIinchRange (Start,kmax//num): GETPSN (num*i,sump+i,product + 1, i) GETPSN (1,1,1,2) ans= SUM (Set (n[2:])) Printans#7587457#running time= 0.266000032425 sdefrun (): Kmin= 2KMax= 12000Res=[]     forKinchRange (kmin,kmax+1): Minn=Getminn (k)ifMinn not inchres:res.append (Minn)PrintSUM (Minn)defGetminn (k): N= k + 1 while(True):ifCheck (n,n,k):returnN N+=1defCheck (prod,sum,k):ifSum<k:returnFalseifProd = = 1:returnsum==kifK==1:returnProd = =sum forDinchRange (2, prod):ifSum-d>=k-1 andprod%d = =0:ifCheck (prod/d,sum-d,k-1):                returnTruereturnFalset0=time.time () run2 () T1=time.time ()Print "running Time=", (T1-T0),"s"            

Project Euler 88:product-sum numbers product and sum