ProjectEuler88：Product-sumnumbers积和数

package Level3;


import java.util.Set;
import java.util.TreeSet;

public class PE088{
    static void run(){
        int Kmin = 2;
        int Kmax = 12000;
        int sum = 0;
        Set set = new TreeSet();
        for(int k=Kmin;k<=Kmax;k++){
            int minN = getMin(k);
            if(set.add(minN))
                sum+=minN;
        }
        System.out.println(sum);
    }
    // 找出k对于最小的n
    static int getMin(int k){
        for(int n=k+1;;n++){
            if(check(n,n,k))
                return n;
        }
    }
    // 一个数拆成成k个数的和或者k个数的积 
    // prod 乘
    // sum 和 
    // 开始的时候这两个数是相等的  都是 prod 或者sum 拆分成k份 
    
//    这里用到的是递归的方法，当 prod2 = prod1 * a ;sum2 = sum1- a 
//            下面就可以检测下一轮了 check(prod2,sum2,k-1)
//    这里用递归也是因为可能出  8 = 2*2*2*1*1 = 2+2+2+1+1  的形式，乘子中有数相同 的情况
//    结束情况： 乘子是1的时候  sum == k k个1的和就是sum了
//     k=1的时候  说明结束了 return prod == sum 
//    下次递归可进行需要：d<= prod  k-1<= sum-d 下面程序很显然的 
    static boolean check(int prod,int sum,int k){
        if(sum return false;
        if(prod == 1) return sum==k;
        if(k==1) return prod ==sum;
        for(int d =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 void main(String[] args){
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");

    }
}

# coding=gbk

import time as time 
def run2():
    kMax = 12000
    n = [2*kMax for i in range(kMax)]

    def getpsn(num,sump,product,start):
        k = num - sump + product
        if k < kMax:
            if num<n[k]:
                n[k] = num
            for i in range(start,kMax//num *2):
                getpsn(num*i,sump+i,product + 1,i)
    getpsn(1,1,1,2)
    ans = sum(set(n[2:]))
    print ans 
# 7587457
# running time= 0.266000032425 s
def run():
    kMin = 2
    kMax = 12000
    res=[]
    for k in range(kMin,kMax+1):
        minN = getMinN(k)
        if minN not in res:
            res.append(minN)
    print sum(minN)
           
def getMinN(k):
    n = k + 1
    while(True):
        if check(n,n,k):
            return n 
        n +=1
        
def check(prod,sum,k):
    if sumreturn False
    if prod == 1:return sum==k 
    if k==1 :return prod ==sum
    for d in range(2,prod):
        if sum-d>=k-1 and prod%d ==0:
            if check(prod/d,sum-d,k-1):
                return True 
    return False

t0 = time.time()
run2() 
t1 = time.time()
print "running time=",(t1-t0),"s"