PAT Basic 1013谁能满分过 给50，必结

[纵骑横飞]

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*1013. 数素数 (20)
 时间限制
 100 ms
 内存限制
 65536 kB
 代码长度限制
 8000 B
 判题程序
 Standard
 作者
 CHEN, Yue

 令Pi表示第i个素数。现任给两个正整数M <= N <= 10^4，请输出PM到PN的所有素数。

 输入格式：

 输入在一行中给出M和N，其间以空格分隔。

 输出格式：

 输出从PM到PN的所有素数，每10个数字占1行，其间以空格分隔，但行末不得有多余空格。
 输入样例：

 5 27

 输出样例：

 11 13 17 19 23 29 31 37 41 43
 47 53 59 61 67 71 73 79 83 89
 97 101 103*/

/*检索素数,放到集合里
 * 遍历器控制输出
 * 改进:检索到第N个素数输出M到N个数
 */
public class Test1013改进 {
	public static void main(String[] args) throws IOException{
		InputStreamReader isr=new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader(isr);
		int one=0;
		int hasRead = br.read();
		int j = 0,M = 0, N = 0;
		while (true) {
			if (hasRead >= '0' && hasRead <= '9') {
				one=one*10+hasRead-48;
				hasRead = br.read();

			} else if (hasRead == ' ') {
				if (j == 0) {
					M =one;
					j++;
				} else{
					N =one;
				}
				hasRead = br.read();
				one = 0;
			} else {
				if (j == 0) {
					M = one;
					j++;
				} else{
					N =one;
				}
				break;
			}

		}
		br.close();

		int Arr[] = new int[N + 1];
		for (int i = 1, k = 0; k <= N; i++) {
			// 判断是不是素数,bool一旦为false直接结束
			boolean bool = true;
			for (j = (int) Math.sqrt(i); bool == true && j > 1; j--) {
				if (i % j == 0)
					bool = false;
			}
			if (bool == true) {
				Arr[k] = i;
				k++;
			}
		}
		j = 1;
		for (int i = M; i <= N; i++, j++) {
			System.out.print(Arr[i]);
			if (j % 10 == 0)
				System.out.println();
			if (i != N && j % 10 != 0) {
				System.out.print(" ");
			}
		}

	}
}

这是我的代码 19分
要满分20分过的(当然用的是Java) 悬赏50

[falsemask_]

这题要用筛选素数的方法，不过不知道为什么我的java代码运行超时，但我之前用c++过了。

[rickylin86]

import java.util.ArrayList;
import java.util.Date;
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class Primes{
	public static void main(String[] args){
		
		System.out.print("Input PM & PN:");
		Scanner keyboard = new Scanner(System.in);
		String content = keyboard.nextLine();
		Date start = new Date();
		String regex = "\\s*(\\d+)\\s+(\\d+)\\s*";
		Matcher matcher = Pattern.compile(regex).matcher(content);
		if(!matcher.matches()){
			System.err.println("Illegal Arguments!");
			System.exit(1);
		}
		int pm = Integer.valueOf(matcher.group(1));
		int pn = Integer.valueOf(matcher.group(2));
		printPrimes(pm,pn);
		Date end = new Date();
		long time = end.getTime() - start.getTime();
		System.out.printf("\ntime:%dms\n",time);
	}

	private static void printPrimes(int startIndex,int endIndex){
		if(startIndex > endIndex || startIndex < 1){
			System.err.println("Illegal Arguments!");
			return;
		}

		ArrayList<Long> primes = new ArrayList<>();
		primes.add(2L);
		primes.add(3L);
		long number = 5;
		long sqrt = 0L;
		long prime = 0L;
		int size = 0;
		for(int count = 2; count < endIndex ; count ++){
			sqrt = (long)(Math.sqrt(number) + 1);
			size = primes.size();
			for(int index = 0;index < size && primes.get(index) < sqrt;){
				if(number % primes.get(index) == 0){
					number += 2;
					index = 0;
					sqrt = (long)(Math.sqrt(number) + 1);
					continue;
				}
				index ++;
			}
			primes.add(number);
			number += 2;
		}

		int lineCount = 0;
		for(int index = startIndex - 1; index < endIndex ; index ++){
			System.out.print(primes.get(index));
			lineCount ++;
			System.out.print(lineCount % 10 == 0 ? "\n" : " ");
		}
		
	}
}

[纵骑横飞]

引用 2 楼 rickylin86 的回复:

import java.util.ArrayList;
import java.util.Date;
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class Primes{
	public static void main(String[] args){
		
		System.out.print("Input PM & PN:");
		Scanner keyboard = new Scanner(System.in);
		String content = keyboard.nextLine();
		Date start = new Date();
		String regex = "\\s*(\\d+)\\s+(\\d+)\\s*";
		Matcher matcher = Pattern.compile(regex).matcher(content);
		if(!matcher.matches()){
			System.err.println("Illegal Arguments!");
			System.exit(1);
		}
		int pm = Integer.valueOf(matcher.group(1));
		int pn = Integer.valueOf(matcher.group(2));
		printPrimes(pm,pn);
		Date end = new Date();
		long time = end.getTime() - start.getTime();
		System.out.printf("\ntime:%dms\n",time);
	}

	private static void printPrimes(int startIndex,int endIndex){
		if(startIndex > endIndex || startIndex < 1){
			System.err.println("Illegal Arguments!");
			return;
		}

		ArrayList<Long> primes = new ArrayList<>();
		primes.add(2L);
		primes.add(3L);
		long number = 5;
		long sqrt = 0L;
		long prime = 0L;
		int size = 0;
		for(int count = 2; count < endIndex ; count ++){
			sqrt = (long)(Math.sqrt(number) + 1);
			size = primes.size();
			for(int index = 0;index < size && primes.get(index) < sqrt;){
				if(number % primes.get(index) == 0){
					number += 2;
					index = 0;
					sqrt = (long)(Math.sqrt(number) + 1);
					continue;
				}
				index ++;
			}
			primes.add(number);
			number += 2;
		}

		int lineCount = 0;
		for(int index = startIndex - 1; index < endIndex ; index ++){
			System.out.print(primes.get(index));
			lineCount ++;
			System.out.print(lineCount % 10 == 0 ? "\n" : " ");
		}
		
	}
}

是BASIC的

[sky_08_06_02]

[纵骑横飞]

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