POJ 刷题系列:1840. Eqs

传送门:1840. Eqs

题意:

等式 a1x31+a2x32+a3x33+a4x34+a5x35=0,其中ai∈[−50,50],i=1,2,3,4,5,xi∈[−50,50],xi≠0,求满足 等式(x1,x2,x3,x4,x5)的解的个数。

思路:
折半搜索,把等式分成两半,左半部分两个变量循环枚举,并用hash记录每个值出现的次数。枚举右半部分是否在hash中出现过,统计个数即可。int会MLE,所以采用char,节省内存。

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/201801/P1840.txt";

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

    static final int maxn = 12500000;

    void read() {
        int a = ni();
        int b = ni();
        int c = ni();
        int d = ni();
        int e = ni();
        char[] hash = new char[maxn * 2];
        for (int i = -50; i <= 50; ++i) {
            if (i == 0) continue;
            for (int j = -50; j <= 50; ++j) {
                if (j == 0) continue;
                int sum = a * i * i * i + b * j * j * j;
                hash[maxn - sum] ++;
            }
        }
        int tot = 0;
        for (int i = -50; i <= 50; ++i) {
            if (i == 0) continue;
            for (int j = -50; j <= 50; ++j) {
                if (j == 0) continue;
                for (int k = -50; k <= 50; ++k) {
                    if (k == 0) continue;
                    int sum = c * i * i * i + d * j * j * j + e * k * k * k;
                    if (sum <= maxn && sum >= -maxn) {
                        tot += hash[sum + maxn];
                    }
                }
            }
        }
        out.println(tot);
    }

    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);
        }
    }
}

POJ 刷题系列:1840. Eqs_搜索