Project Euler 77：Prime summations

原题：

Prime summations

It is possible to write ten as the sum of primes in exactly five different ways:

7 + 3
5 + 5
5 + 3 + 2
3 + 3 + 2 + 2
2 + 2 + 2 + 2 + 2

What is the first value which can be written as the sum of primes in over five thousand different ways?

翻译：

素数加和

将10写成素数的和有5种不同的方式：

7 + 3
5 + 5
5 + 3 + 2
3 + 3 + 2 + 2
2 + 2 + 2 + 2 + 2

写成素数的和有超过五千种不同的方式的数最小是多少？

思路：

动态规划题目

我直接网上找的代码

但是大家写的好多都一样的

附：之前的动态规划介绍

Java程序：

package Level3;

import java.util.ArrayList;
import java.util.Iterator;

public class PE077 {

    void run(){
        int limit = 5000;
        dp(limit);
    }

    void dp(int limit){
        ArrayList<Integer> plist = listPrime(limit/5);
        int target = 2;
        while(true){
        int[] ways = new int[target+1];
        ways[0] = 1;
        for(int i=0;i<plist.size();i++){
            for(int j=(int) plist.get(i);j<=target;j++)
                ways[j] += ways[j-(int) plist.get(i)];
        }
//        System.out.println(target+" " + ways[target]);
        if(ways[target] > limit) break;
        target++;
        }
        System.out.println(target);

    }
//    71
//    running time=0s7ms
    ArrayList<Integer> listPrime(int limit){
        int prime[] = new int[limit];
        ArrayList<Integer> plist = new ArrayList<Integer>();
        boolean isPrime = true;
        prime[0]=2;
        plist.add(2);
        int p=1;
        for(int i=2;i<limit;i++){
            isPrime = true;
                Iterator<Integer> it = plist.iterator();
                while(it.hasNext() &&isPrime){
                    int prm=it.next();
                    if(i%prm==0){// 说明 i 不是素数
                        isPrime = false;
                        break;
                    }
                }

            if(isPrime==true)
                plist.add(i);
        }

        return plist;

    }

    public static void main(String[] args) {
        long t0 = System.currentTimeMillis();
        new PE077().run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");
    }

}

Python程序：

import time 

def sieve(limit):
    primes= []
    is_prime = True
    primes.append(2)
    for i in range(3,limit):
        is_prime = True
        for ps in primes:
            if i%ps ==0:
                is_prime = False
                break
        if is_prime==True:
            primes.append(i)
    return primes

def dp(limit):
    primes = sieve(1000)
    target = 2
    while True:
        ways = [1] + [0]*target
        for prime in primes:
            for j in range(prime,target+1):
                ways[j] += ways[j-prime]
        if ways[target]> limit:
            break
        target+=1
    print target
#     71
# running time 0.00999999046326 s
if __name__==‘__main__‘:
    t0 = time.time()
    limit = 5000
    dp(limit)
    print"running time",(time.time() - t0),"s"