用sql语句查询For every match where 'GER' scored, show matchid, match date and the number of goals scored by 'GER'

以下是查询语句：

SELECT matchid, date, COUNT(*) AS goals_scored
FROM match
JOIN goal ON match.id = goal.matchid
WHERE team1 = 'GER' AND goal.teamid = 'GER'
GROUP BY matchid, date;

这条 SQL 语句首先使用 JOIN 将 match 表和 goal 表连接，然后使用 WHERE 子句筛选出所有 team1 为 'GER' 且 goal 表中 teamid 为 'GER' 的比赛。最后使用 GROUP BY 对每场比赛的 matchid 和 date 进行分组，并使用 COUNT(*) 计算出 'GER' 这个队在每场比赛中进球的数量。

相关问题

(Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

Here's a Python program that implements the algorithm described in the problem statement:

import math

def smallest_perfect_square_factor(m):
    factors = []
    # Find all smallest factors of m
    while m % 2 == 0:
        factors.append(2)
        m = m // 2
    for i in range(3, int(math.sqrt(m))+1, 2):
        while m % i == 0:
            factors.append(i)
            m = m // i
    if m > 2:
        factors.append(m)
    # Compute n as the product of factors that appear an odd number of times
    odd_factors = [x for x in factors if factors.count(x) % 2 == 1]
    n = 1
    for f in odd_factors:
        n *= f
    return n

m = int(input("Enter an integer m: "))
n = smallest_perfect_square_factor(m)
print(f"The smallest integer n such that m * n is a perfect square is {n}.")

The program first computes all the smallest factors of `m` and stores them in an array list called `factors`. Then, it identifies the factors that appear an odd number of times in the list, and multiplies them together to obtain `n`. Finally, the program prints out the value of `n`.

Java (Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

Here's the Java code to solve the problem:

import java.util.ArrayList;
import java.util.Scanner;

public class PerfectSquare {
   public static void main(String[] args) {
      Scanner input = new Scanner(System.in);
      System.out.print("Enter an integer m: ");
      int m = input.nextInt();
      ArrayList<Integer> factors = new ArrayList<Integer>();
      
      int factor = 2;
      while (factor <= m) {
         if (m % factor == 0) {
            factors.add(factor);
            m /= factor;
         }
         else {
            factor++;
         }
      }
      
      int n = 1;
      for (int i = 0; i < factors.size(); i++) {
         int count = 1;
         for (int j = i + 1; j < factors.size(); j++) {
            if (factors.get(i) == factors.get(j)) {
               count++;
            }
         }
         if (count % 2 != 0) {
            n *= factors.get(i);
         }
      }
      
      System.out.println("The smallest integer n such that m * n is a perfect square is: " + n);
   }
}

Explanation:

1. We first prompt the user to enter an integer `m` using the `Scanner` class.
2. We then create an `ArrayList` called `factors` to store all the smallest factors of `m`.
3. We use a `while` loop to find all the smallest factors of `m`. If `m` is divisible by `factor`, we add it to the `factors` list and divide `m` by `factor`. Otherwise, we increment `factor` by 1.
4. We then use two nested `for` loops to check how many times each factor appears in the `factors` list. If a factor appears an odd number of times, we multiply it with `n`.
5. Finally, we print out the value of `n`.