UVa1210 - Sum of Consecutive Prime Numbers(欧拉筛法即线性筛法)

本文深入探讨了信息技术领域的多个细分技术领域,包括前端开发、后端开发、移动开发等,通过详细解析每个领域的重要技术和实践,为读者提供了一个全面的技术视野。同时,文章还涉及大数据开发、开发工具、嵌入式硬件、音视频基础、AI音视频处理等多个关键领域,旨在帮助开发者深入了解技术生态,促进创新与应用。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2 + 3 + 5 + 7 + 11 + 13, 11 + 13 + 17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. Your mission is to write a program that reports the number of representations for the given positive integer.

Input 

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10000, inclusive. The end of the input is indicated by a zero.

Output 

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

Sample Input 

2 
3 
17 
41
20 
666 
12 
53 
0

Sample Output 

1 
1 
2 
3 
0 
0 
1 
2
用欧拉筛法(即线性筛法)求出10000内的素数,然后用数组f[i]表示从第1个素数到第i个素数的和,用两重循环判断f[i] - f[j]是否等于n

import java.io.FileInputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.OutputStreamWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;

public class Main implements Runnable
{
	private static final boolean DEBUG = false;
	private static final int N = 10001;
	private BufferedReader cin;
	private PrintWriter cout;
	private StreamTokenizer tokenizer;
	private boolean[] vis;
	private ArrayList<Integer> vPrime = new ArrayList<Integer>();
	private int[] f;
	private int n;
	
	private void preprocess()
	{
		vis = new boolean[N];
		vis[0] = vis[1] = true;
		
		for (int i = 2; i < N; i++) {
			if (!vis[i]) {
				vPrime.add(i);
			}
			
			int size = vPrime.size(); 
			for (int j = 0; j < size && i * vPrime.get(j) < N; j++) {
				int num = vPrime.get(j);
				vis[i * num] = true;
				
				if (i % num == 0) break;
			}
		}
		
		int size = vPrime.size();
		f = new int[size + 1];
		
		f[0] = 0;
		for (int i = 0; i < size; i++) {
			f[i + 1] = f[i] + vPrime.get(i);
		}
	}
	
	private void init() 
	{
		try {
			if (DEBUG) {
				cin = new BufferedReader(new InputStreamReader(
						new FileInputStream("d:\\OJ\\uva_in.txt")));
			} else {
				cin = new BufferedReader(new InputStreamReader(System.in));
			}

			tokenizer = new StreamTokenizer(cin);
			
			cout = new PrintWriter(new OutputStreamWriter(System.out));
			preprocess();
			
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	private String next() 
	{
		try {
			 tokenizer.nextToken(); 
			 if (tokenizer.ttype == StreamTokenizer.TT_EOF) return null; 
			 else if (tokenizer.ttype == StreamTokenizer.TT_NUMBER) { 
			 	return String.valueOf((int)tokenizer.nval); 
			} else return tokenizer.sval;
		} catch (Exception e) {
			e.printStackTrace();
			return null;
		}
	}

	private boolean input() 
	{
		n = Integer.parseInt(next());
		if (n == 0) return false;
		
		return true;
	}
	
	private void solve() 
	{
		int size = f.length;
		
		int ans = 0;
		for (int i = 0; i < size; i++) {
			for (int j = i + 1; j < size; j++) {
				if (f[j] - f[i] == n) ans++;
			}
		}
		
		cout.println(ans);
		cout.flush();
	}

	public void run()
	{
		init();
		while (input()) {
			solve();
		}
	}
	
	public static void main(String[] args) 
	{
		new Thread(new Main()).start();
	}
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

kgduu

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值