挑战程序竞赛系列（40）：4.1模运算的世界（3）

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

• POJ 2115: C Looooops

POJ 2115: C Looooops

题目的Loooo…有点长啊，哈哈。

此题的思路很简单，只要列出一个式子即可，如下：

$$(A + CX) \equiv B \mod \space 2^k$$

整理一下，即求同余表达式$CX \equiv (B - A) \mod \space 2^k$

那么问题就转换成同余表达式如何求解了？可以参考《挑战》P291关于逆元的介绍，这里给出自己的求解思路。

首先，同余表达式可能有解，也可能无解，就拿式子$ax \equiv b \mod m$而言，我们可以写成如下表达式：$ax - my = b$

所以为了该方程有解，必须满足d = gcd(a,m)， b能够被d整除，否则等式左边将出现小数，一定无解。

既然判断了同余表达式是否有解，接下来就可以化为：

$$d = gcd(a, m) \\ (a / d) x \equiv(b / d) \mod m / d$$

此时，因为$gcd(a/d, m/d) = 1$，满足$ax \equiv 1 \mod m$关于逆元的定义条件，一定存在解，我们可以求出逆元$x_0 = (a / d)^{-1}$

于是可以表示为：

$$(a / d) x \equiv(b / d) \mod m / d \\ x \equiv (a/d)^{-1}(b/d) \mod m / d$$

对应的代码如下：

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.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201708/P2115.txt";

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

    void solve() {
        while (true){
            int A = ni();
            int B = ni();
            int C = ni();
            int K = ni();
            if (A + B + C + K == 0) break;
            long ans = modular_linear_equation_solver(C, B - A, (long)1 << K);
            if (ans == -1){
                out.println("FOREVER");
            }
            else{
                out.println(ans);
            }
        }
    }

    class Pair{
        long d;
        long x;
        long y;
        // a * x + b * y = d
        public Pair(long d, long x, long y){
            this.d = d;
            this.x = x;
            this.y = y;
        }
    }

    public Pair extgcd(long a, long b){ // a > b or a < b
        if (b == 0){
            return new Pair(a, 1, 0);
        }
        else{
            Pair p = extgcd(b, a % b);
            Pair ans = new Pair(0, 0, 0);
            ans.d = p.d;
            ans.x = p.y;
            ans.y = p.x - (a / b) * p.y;
            return ans;
        }
    }

    public long mod_inverse(long a, long m){
        Pair p = extgcd(a, m);
        if (p.d != 1) return -1;
        return (p.x % m + m) % m;
    }

    public long modular_linear_equation_solver(long a, long b, long m){
        Pair p = extgcd(a, m);
        if (b % p.d != 0) return -1;
        long x0 = mod_inverse(a / p.d, m / p.d);
        return (x0 * (b / p.d) % (m / p.d) + (m / p.d)) % (m / p.d);
    }



    FastScanner in;
    PrintWriter out;

    void run() throws IOException {
        boolean oj;
        try {
            oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
        } 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();
        solve();
        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);
        }
    }
}

注意：逆元和同余方程的解都有可能为负，所以需要加mod处理，注意下。