Codeforces Round #140Div2

What do I turn?

很简单，运用叉积即可得到答案。

package com.wangzhu.codeforces;

//What do I turn?
import java.io.BufferedInputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Round140Div2 {

    public static void main(String[] args) throws FileNotFoundException {
        System.setIn(new FileInputStream("data.in"));
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        int Len = 3;
        Point[] points = new Point[Len];
        long temp = -1;
        while (cin.hasNext()) {
            for (int i = 0; i < Len; i++) {
                points[i] = new Point(cin.nextLong(), cin.nextLong());
            }
            temp = xmult(points[0], points[1], points[2]);
            if (temp < 0) {
                System.out.println("RIGHT");
            } else if (temp > 0) {
                System.out.println("LEFT");
            } else {
                System.out.println("TOWARDS");
            }
        }
        cin.close();
    }

    public static long xmult(Point p1, Point p2, Point p3) {
        return (p1.x - p2.x) * (p2.y - p3.y) - (p1.y - p2.y) * (p2.x - p3.x);
    }
}

class Point {
    long x, y;

    public Point(long x, long y) {
        this.x = x;
        this.y = y;
    }
}

Anniversary

这题我知道gcd(a,b)=gcd(fib(a),fib(b))

接下来问别人的！

首先fibonacci数具有性质  gcd(f(i),f(j))=f(gcd(i,j))，所以等于就要找l到r之间的k个数他们最大公约数最大，我们设找到的最大公约数是p。

也就是要球最大的p使得[r/p]-[(l-1)/p]>=k 其中[x]是取整的意思。

最大的p满足条件说明p满足，p+1就不满足了（这个题目p不具有单调性，也有情况p不满足p+1满足的，但是我们要求的是最大值，所以一定是p满足条件比p大的都不满足条件），然后显然我们要求的最大的p满足[r/(p+1)]<[r/p]，因为随着p的增大-[(l-1)/p]只可能增大，所以如果p满足条件p+1不满足的话一定是第一项发生了缩小。所以我们只要找[r/(p+1)]<[r/p]的p就可以了。这个p的话是1，2，3，...，[sqrt(r)]，[r/[sqrt(r)]], [r/[sqrt(r)]-1], ... ,[r/2], [r] 一共有2sqrt(r)种情况，然后从大向小枚举就可以了，然后f(p)就是最后答案，这个用快速幂加速一下就可以了。

package com.wangzhu.codeforces;

//Anniversary
import java.io.BufferedInputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Arrays;
import java.util.Scanner;

public class Round140Div2 {

    public static void main(String[] args) throws FileNotFoundException {
        System.setIn(new FileInputStream("data.in"));
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        long m = -1, L = -1, R = -1, k = -1, n = -1;
        int len = -1, temp = -1;
        long[] arr = null;
        while (cin.hasNext()) {
            m = cin.nextLong();
            L = cin.nextLong();
            R = cin.nextLong();
            k = cin.nextLong();
            temp = (int) Math.sqrt(1.0 * R);
            len = temp * 2;
            arr = new long[len];
            for (int i = 1; i <= temp; i++) {
                arr[i - 1] = i;
            }
            for (int i = 0, j = temp; i < temp; i++) {
                arr[j++] = R / arr[i];
            }
            Arrays.sort(arr);
            for (int i = len - 1; i >= 0; i--) {
                if ((R / arr[i] - (L - 1) / arr[i] >= k)) {
                    n = arr[i];
                    break;
                }
            }
            cal(n, m);
        }
        cin.close();
    }

    private static void cal(long k, long MOD) {
        long[][] numa = new long[2][2];
        long[][] numb = new long[2][2];
        numa[0][0] = 1;
        numa[0][1] = 0;
        numa[1][0] = 0;
        numa[1][1] = 1;
        numb[0][0] = 0;
        numb[0][1] = 1;
        numb[1][0] = 1;
        numb[1][1] = 1;
        while (k != 0) {
            if (k % 2 == 1) {
                numa = matrix(numa, numb, MOD);
            }
            numb = matrix(numb, numb, MOD);
            k >>= 1;
        }
        System.out.println(numa[0][1]);
    }

    private static long[][] matrix(long[][] a, long[][] b, long MOD) {
        long[][] temp = new long[2][2];
        temp[0][0] = (a[0][0] * b[0][0] % MOD + a[0][1] * b[1][0] % MOD) % MOD;
        temp[0][1] = (a[0][0] * b[0][1] % MOD + a[0][1] * b[1][1] % MOD) % MOD;
        temp[1][0] = (a[1][0] * b[0][0] % MOD + a[1][1] * b[1][0] % MOD) % MOD;
        temp[1][1] = (a[1][0] * b[0][1] % MOD + a[1][1] * b[1][1] % MOD) % MOD;
        return temp;
    }

}