A. Olympic Ranking

按照题意对三个数进行排序就行，我是用的一个int数组来排序，int数组的最后一个参数是初始位置。
把国家名字按照输入顺序存储，和上面int[3]是一样的，便于int排序后找到国家名。

踩坑：输入的时候先输入三个int，再输入带空格的string会读不进来，不知道为什么，所以只能直接读一整行然后split了

自定义排序

/**
 * @author :Changersh
 * @date : 2023/4/29 13:04
 */
 
import java.io.*;
import java.util.*;
import java.lang.*;
 
public class Main {
    private static int T, N = 310, n;
    private static String[] s = new String[N];
    private static Integer[][] a;
 
    public static void main(String[] args) {
        n = sc.nextInt();
        a = new Integer[n][4];
        for (int i = 0; i < n; i++) {
            String[] sp = sc.nextLine().split(" ");
            a[i][0] = Integer.parseInt(sp[0]);
            a[i][1] = Integer.parseInt(sp[1]);
            a[i][2] = Integer.parseInt(sp[2]);
            a[i][3] = i;
            s[i] = "";
            for (int j = 3; j < sp.length - 1; j++) {
                s[i] += sp[j] + " ";
            }
            s[i] += sp[sp.length - 1];
        }
 
        Arrays.sort(a, new Comparator<Integer[]>() {
            @Override
            public int compare(Integer[] x, Integer[] y) {
                if (!x[0].equals(y[0])) return y[0] - x[0];
                if (!x[1].equals(y[1])) return y[1] - x[1];
                return y[2] - x[2];
            }
        });
 
        pw.println(s[a[0][3]]);
 
        pw.close();
    }
}

B. Aliquot Sum

给了8秒，暴力求就行
sum一般初始化为0，答案有 1 的时候 i 一般也从 1 开始，别改。

约数和

/**
 * @author :Changersh
 * @date : 2023/5/1 8:59
 */
 
import java.io.*;
import java.util.*;
import java.lang.*;
 
public class Main {
    private static int T, n, sum;
 
    public static void main(String[] args) {
        T = sc.nextInt();
        while (T-- > 0) solve();
 
        pw.close();
    }
 
    private static void solve() {
        n = sc.nextInt();
        sum = 0;
        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                sum += i;
                if (n / i != i) sum += (n / i);
            }
        }
        sum -= n;
        if (sum > n) pw.println("abundant");
        else if (sum < n) pw.println("deficient");
        else pw.println("perfect");
    }
}

C. A Sorting Problem

由题意得，求逆序对

求逆序对

/**
 * @author :Changersh
 * @date : 2023/4/29 13:32
 */

import java.io.*;
import java.util.*;
import java.lang.*;

public class Main {
    private static int N = 100005, n;
    private static int[] a, cnt;
    private static long ans = 0;

    public static void main(String[] args) {
        n = sc.nextInt();
        a = new int[n];
        cnt = new int[n];
        for (int i = 0; i < n; i++) a[i] = sc.nextInt();
        merge_sort(0, n - 1);

        pw.println(ans);

        pw.close();
    }

    private static void merge_sort(int l, int r) {
        if (l >= r) return;
        int mid = l + ((r - l) >> 1);
        merge_sort(l, mid);
        merge_sort(mid + 1, r);

        // merge
        int i = l, j = mid + 1, p = l;
        while (i <= mid && j <= r) {
            if (a[i] <= a[j]) cnt[p++] = a[i++];
            else {
                ans += mid - i + 1;
                cnt[p++] = a[j++];
            }
        }
        while (i <= mid) cnt[p++] = a[i++];
        while (j <= r) cnt[p++] = a[j++];

        for (int k = l; k <= r; k++) a[k] = cnt[k];
    }
}

D. Drunk Passenger

额，猜吧

数学

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException{
        BufferedReader re = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer r = new StreamTokenizer(re);
        PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));

        r.nextToken();
        int n = (int)r.nval;

        double ans = 1;
        if (n != 2)
            ans = 1.0/(n-1) + (double)(n-2)/(n-1)/2 ;
        pr.println(ans);
        pr.flush();
    }
}

E. Eatcoin

强烈建议用python，试了一下10^100都没超，太厉害了，不用考虑溢出了
1 ^ 5 + 2 ^ 5 + 3 ^ 5+......+ n ^ 5 = (2 * (n + 1) ^ 3 * n * n * n - (n + 1) ^ 2 * n * n) // 12
然后二分天数即可

二分+公式

s = input().split(" ")
p = int(s[0])
q = int(s[1])

x = 0
y = 0

sum = 0
for i in range(1, 20000):
    sum += p
    sum -= q * (i ** 5)
    if (sum <= 0):
        x += p
        break
    x = sum
print(x)


l = 1
r = 10 ** 100
ans = 0
while l <= r:
    mid = l + (r - l) // 2
    sum = q * (mid * mid) * (mid + 1) * (mid + 1) * (2 * mid * mid + 2 * mid - 1) // 12 - p * mid + x
    if (sum >= 10**99):
        r = mid - 1
        ans = mid
    else:
        l = mid + 1

print(ans)

J. JavaScript

队友写的，我没写
看了一下，是个模拟，给你两个字符串，包含字母或者数字。
如果包含字母就不能转化成数字了，结果是 NaN。
两个都是纯数字组成的情况下，才能计算，计算 (x - y)

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException{
        BufferedReader re = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer r = new StreamTokenizer(re);
        PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));

        String[] num = re.readLine().split(" ");
        boolean flag1 = false,flag2 = false;
        for (int i = 0; i < num[0].length(); i++) {
            if(num[0].charAt(i) > '9' || num[0].charAt(i) < '0'){
                flag1 = true;
                break;
            }
        }
        for (int i = 0; i < num[1].length(); i++) {
            if(num[1].charAt(i) > '9' || num[1].charAt(i) < '0'){
                flag2 = true;
                break;
            }

        }
        if( flag1 && flag1){
            pr.println("NAN");
        }
        else {
            pr.println(Integer.parseInt(num[0]) - Integer.parseInt(num[1]));
        }
        pr.flush();
    }
}