AcWing 730. 机器人跳跃问题（二分）

Problem

初看题是看不出要用二分的，但是题目要求找出符合条件最小的E，这个就是符合二分的条件，最值，单调性，因为所有大于E的值都可成立。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);
    static int N = 100010, n;
    static int a[] = new int[N];

    public static void main(String[] args) throws Exception {

        n = Integer.parseInt(br.readLine());
        String s[] = br.readLine().split(" ");
        for (int i = 1; i <= n; i++) a[i] = Integer.parseInt(s[i - 1]);

        int l = 0, r = N;
        while (l < r) {
            int mid = l + r >> 1;
            if (check(mid)) r = mid;
            else l = mid + 1;
        }
        pw.println(r);

        pw.flush();
        pw.close();
        br.close();
    }

    public static boolean check(int e) {
        for (int i = 1; i <= n; i++) {
            e = e * 2 - a[i];
            if (e > N) return true;
            if (e < 0) return false;
        }
        return true;
    }

}