POJ 刷题系列:3006. Dirichlet's Theorem on Arithmetic Progressions

来源:互联网 发布:如何搜索二次元软件 编辑:程序博客网 时间:2024/06/03 17:22

POJ 刷题系列:3006. Dirichlet’s Theorem on Arithmetic Progressions

传送门:3006. Dirichlet’s Theorem on Arithmetic Progressions

题意:

给出一个递增序列,求该递增序列中第n个素数。

思路:
这些水题怎么都跟素数有关。。。实际上给定初项a和公差d,以及可能的素数,能判断该素数是否在这个递增数列中。

if (prime - a) % d == 0 说明是递增数列中的一个元素所以,只要遇到素数属于递增数列,计数,直到第n个就输出答案。

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/P3006.txt";    public static void main(String[] args) throws IOException {        new Main().run();    }    static final int MAX_N = 1000000 + 16;    boolean[] isPrime;    int[] primes;    int tot;    void seive() {        isPrime = new boolean[MAX_N];        primes  = new int[MAX_N];        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;                }            }        }    }    int solve(int a, int d, int n) {        int ans = 0;        for (int j = 0, i = 0; j < n; ++j, ++i) {            while (primes[i] - a < 0 || (primes[i] - a) % d != 0) {                i ++;            }            ans = primes[i];        }        return ans;    }    void read() {        seive();        while (true) {            int a = ni();            int d = ni();            int n = ni();            if (a + d + n == 0) break;            out.println(solve(a, d, n));        }    }    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);        }    }}

这里写图片描述

阅读全文
0 0
原创粉丝点击