POJ 刷题系列:2262. Goldbach's Conjecture

POJ 刷题系列:2262. Goldbach’s Conjecture

传送门:POJ 2262. Goldbach’s Conjecture

题意:

给定一个大于4的数num,求两个奇素数使得num = p1 + p2.

思路: 打一个素数表,枚举小于num的素数p1,接着二分查找num-p2是否在素数表中,有则输出答案。

代码如下:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201712/P2262.txt";

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    static final int MAX_N = 1000000 + 16;
    boolean[] isPrime = new boolean[MAX_N];
    int[] primes = new int[MAX_N];
    int tot;

    void seive() {
        Arrays.fill(isPrime, true);
        for (int i = 2; i < MAX_N; ++i) {
            if (isPrime[i]) {
                primes[tot++] = i;
                for (int j = 2 * i; j < MAX_N; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    void solve(int num) {

        for (int i = 0; i < tot && primes[i] < num; ++i) {
            int p1 = primes[i];
            int idx = Arrays.binarySearch(primes, 0, tot, num - p1);
            if (idx >= 0) {
                out.println(num + " = " + p1 + " + " + (num - p1));
                return;
            }
        }
        out.println("Goldbach's conjecture is wrong.");
    }

    void read() {
        seive();
        while (true) {
            int num = ni();
            if (num == 0) break;
            solve(num);
        }
    }

    FastScanner in;
    PrintWriter out;

    void run() throws IOException {
        boolean oj;
        try {
            oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");
        } catch (Exception e) {
            oj = System.getProperty("ONLINE_JUDGE") != null;
        }

        InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
        in = new FastScanner(is);
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        read();
        out.flush();
        if (!oj){
            System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
        }
    }

    public boolean more(){
        return in.hasNext();
    }

    public int ni(){
        return in.nextInt();
    }

    public long nl(){
        return in.nextLong();
    }

    public double nd(){
        return in.nextDouble();
    }

    public String ns(){
        return in.nextString();
    }

    public char nc(){
        return in.nextChar();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        boolean hasNext;

        public FastScanner(InputStream is) throws IOException {
            br = new BufferedReader(new InputStreamReader(is));
            hasNext = true;
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    hasNext = false;
                    return "##";
                }
            }
            return st.nextToken();
        }

        String next = null;
        public boolean hasNext(){
            next = nextToken();
            return hasNext;
        }

        public int nextInt() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Integer.parseInt(more);
        }

        public long nextLong() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Long.parseLong(more);
        }

        public double nextDouble() {
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return Double.parseDouble(more);
        }

        public String nextString(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more;
        }

        public char nextChar(){
            if (next == null){
                hasNext();
            }
            String more = next;
            next = null;
            return more.charAt(0);
        }
    }

    static class D{

        public static void pp(int[][] board, int row, int col) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < row; ++i) {
                for (int j = 0; j < col; ++j) {
                    sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
                }
            }
            System.out.println(sb.toString());
        }

        public static void pp(char[][] board, int row, int col) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < row; ++i) {
                for (int j = 0; j < col; ++j) {
                    sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
                }
            }
            System.out.println(sb.toString());
        }
    }

    static class ArrayUtils {

        public static void fill(int[][] f, int value) {
            for (int i = 0; i < f.length; ++i) {
                Arrays.fill(f[i], value);
            }
        }

        public static void fill(int[][][] f, int value) {
            for (int i = 0; i < f.length; ++i) {
                fill(f[i], value);
            }
        }

        public static void fill(int[][][][] f, int value) {
            for (int i = 0; i < f.length; ++i) {
                fill(f[i], value);
            }
        }
    }

    static class Num{
        public static <K> void inc(Map<K, Integer> mem, K k) {
            if (!mem.containsKey(k)) mem.put(k, 0);
            mem.put(k, mem.get(k) + 1);
        }
    }
}

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

来自专栏机器学习入门

挑战程序竞赛系列(95):3.6数值积分(1)

挑战程序竞赛系列(95):3.6数值积分(1) 传送门:AOJ 1313: Intersection of Two Prisms 题意: 有一个侧棱与Z轴平行...

19210
来自专栏Hongten

Hibernate 过滤器

通过调用Session对象的setFilter()和enableFilter()方法使用过滤器。

442
来自专栏吴伟祥

java解析任意层的json数据(递归解析) 原

641
来自专栏机器学习入门

POJ 刷题系列:2109. Power of Cryptography

题意: 给定n,p,求k,使得kn=pk^n = p 思路: 这不应该放在贪心里啊!!!刷新了我对double的认识,实际上double的表示范围是巨大的...

1765
来自专栏机器学习入门

POJ 刷题系列:2255. Tree Recovery

POJ 刷题系列:2255. Tree Recovery 传送门:POJ 2255. Tree Recovery 题意: 给出先序和中序,求后序。 思路: ...

1977
来自专栏机器学习入门

挑战程序竞赛系列(70):4.7后缀数组(2)

挑战程序竞赛系列(70):4.7后缀数组(2) 传送门:POJ 1509: Glass Beads 题意: The description of the ne...

1967
来自专栏机器学习入门

挑战程序竞赛系列(78):4.3 2-SAT(2)

挑战程序竞赛系列(78):4.3 2-SAT(2) 传送门:POJ 3678: Katu Puzzle 题意: 某组合电路有N个输入,M个与或异或门将其两两...

1966
来自专栏机器学习入门

挑战程序竞赛系列(91):3.6凸包(2)

挑战程序竞赛系列(91):3.6凸包(2) 传送门:POJ 1113: Wall 题意参考hankcs: http://www.hankcs.com/pro...

1926
来自专栏小樱的经验随笔

Codeforces 777B Game of Credit Cards

B. Game of Credit Cards time limit per test:2 seconds memory limit per test:256 ...

2765
来自专栏机器学习入门

挑战程序竞赛系列(94):3.6凸包(5)

挑战程序竞赛系列(94):3.6凸包(5) 传送门:POJ 2079: Triangle 题意: 求三个点构成的最大三角形面积。 思路: 可以证明,三点构...

1669

扫码关注云+社区