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社区首页 >专栏 >挑战程序竞赛系列(42):4.1模运算的世界(4)

挑战程序竞赛系列(42):4.1模运算的世界(4)

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发布2019-05-26 09:39:32
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发布2019-05-26 09:39:32
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文章被收录于专栏:机器学习入门

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://cloud.tencent.com/developer/article/1434704

挑战程序竞赛系列(42):4.1模运算的世界(4)

详细代码可以fork下Github上leetcode项目,不定期更新。

练习题如下:

POJ 2720: Last Digits

此题的难点在于如何解决溢出问题,这里用到数论的一个知识点,没办法只能照葫芦画瓢。

具体利用了欧拉函数,算法如下:

代码语言:javascript
复制
    public long f(int b, int i, int mod){
        if (i == 0) return 1;
        else if (mod == 1) return 0;
        else if (fbx[b][i] < 0){
            int euler = (int) euler_phi(mod);
            return pow(b, f(b, i - 1, euler) + euler, mod);
        }
        else{
            return fbx[b][i] % mod;
        }
    }

好了,f函数编写完成之后,我们用快速幂来实现pow函数,先打表fbx把能算的算出来。

最后再用一个cache把已经算出的值缓存起来,因为后续输出变动的原因在于n的变化。

代码如下:

代码语言:javascript
复制
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.StringTokenizer;

public class Main{

    String INPUT = "./data/judge/201708/P2720.txt";

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

    static final int MAX_N = 101;
    int[][] fbx = new int[MAX_N][MAX_N];
    int[] ten_power = new int[8];

    public int power_of(int x, int y){
        if (y < 0) return -1;
        int res = 1;
        while (y > 0){
            if ((y & 1) != 0){
                res = res * x;
                if (res >= ten_power[7] || res <= 0) return -1;
            }
            y >>= 1;
            x = x * x;
        }
        return res;
    }

    public void prepare(){
        ten_power[0] = 1;
        for (int i = 1; i < 8; ++i){
            ten_power[i] = ten_power[i - 1] * 10;
        }

        for (int i = 0; i < MAX_N; ++i) Arrays.fill(fbx[i], -1);
        for (int i = 1; i < MAX_N; ++i){
            fbx[i][0] = 1;
            for (int j = 1; j < MAX_N; ++j){
                fbx[i][j] = power_of(i, fbx[i][j - 1]);
            }
        }
    }


    int[][] cache = new int[MAX_N][MAX_N];
    void solve() {

        prepare();
        for (int i = 0; i < MAX_N; ++i) Arrays.fill(cache[i], -1);

        while (true){
            int b = ni();
            if (b == 0) break;
            int i = ni();
            int n = ni();
            int ans = 0;
            if (cache[b][i] < 0){
                if (b == 1){
                    ans = 1;
                }
                else ans = (int) f(b, i, ten_power[7]);
                cache[b][i] = ans;
            }
            ans = cache[b][i] % ten_power[n];

            StringBuilder sb = new StringBuilder();
            for (int j = 0; j < n - String.valueOf(ans).length(); ++j){
                sb.append("0");
            }
            out.println(sb.toString() + ans);
        }
    }

    public long euler_phi(int m){
        long res = m;
        for (int i = 2; i <= m / i; ++i){
            if (m % i == 0){
                res = res / i * (i - 1);
                for (; m % i == 0; m /= i);
            }
        }
        if (m != 1){
            res = res / m * (m - 1);
        }
        return res;
    }

    public long f(int b, int i, int mod){
        if (i == 0) return 1;
        else if (mod == 1) return 0;
        else if (fbx[b][i] < 0){
            int euler = (int) euler_phi(mod);
            return pow(b, f(b, i - 1, euler) + euler, mod);
        }
        else{
            return fbx[b][i] % mod;
        }
    }

    public long pow(long base, long n, long mod){
        long res = 1;
        while (n > 0){
            if ((n & 1) != 0){
                res = res * base % mod;
            }
            n >>= 1;
            base = base * base % mod;
        }
        return res;
    }

    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);
        }
    }
}
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原始发表:2017年08月30日,如有侵权请联系 cloudcommunity@tencent.com 删除

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