### Product-sum numbers题号：88 难度： 40 中英对照

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.

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

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 are as follows.

k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 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 only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2≤k≤12000?

### Code


public final class p88 {
public static void main(String[] args) {
long start=System.nanoTime();
String result = run();
long end=System.nanoTime();
System.out.println(result);
System.out.println( (end-start)/1000000 + "ms" );
}
static public boolean vis[]=new boolean[24444];
static public int psn[]=new int[12222];

static public String run(){
int ans=0;
getPSN(1,1,1);
for(int k=2;k<=12000;k++){
if(vis[psn[k]]==false){
vis[psn[k]]=true;
ans+=psn[k];
}
}
return Integer.toString(ans);
}
static public void getPSN(int n,int psum,int pcnt){
//n是当前的数字，psum是其分解的因子之和，pcnt是分解的因子个数
int k=n-psum+pcnt;//需要添加1的个数
if(k>12000) return;
if(psn[k]==0 || psn[k]>n)
psn[k]=n;
for(int i=2;i*n<=24000;i++)
getPSN(i*n,psum+i,pcnt+1);//计算i*n的psn
}
}
7587457
48ms