解题思路

这是一个数列求和问题。需要计算数列前 项的和,其中每一项是前一项的平方根。

关键点:

  1. 每一项是前一项的平方根
  2. 需要保留2位小数
  3. 处理多组输入
  4. 使用 类型避免精度损失

算法步骤:

  1. 读取每组输入的
  2. 计算数列的每一项
  3. 累加求和
  4. 格式化输出结果

代码

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    double calculateSum(int n, int m) {
        double sum = 0;
        double current = n;
        
        // 计算前m项的和
        for (int i = 0; i < m; i++) {
            sum += current;
            current = sqrt(current);
        }
        
        return sum;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, m;
    Solution solution;
    
    // 处理多组输入
    while (cin >> n >> m) {
        double result = solution.calculateSum(n, m);
        // 设置输出格式
        cout << fixed << setprecision(2) << result << endl;
    }
    
    return 0;
}
import java.util.*;

class Solution {
    public double calculateSum(int n, int m) {
        double sum = 0;
        double current = n;
        
        // 计算前m项的和
        for (int i = 0; i < m; i++) {
            sum += current;
            current = Math.sqrt(current);
        }
        
        return sum;
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        Solution solution = new Solution();
        
        // 处理多组输入
        while (sc.hasNext()) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            
            double result = solution.calculateSum(n, m);
            // 格式化输出
            System.out.printf("%.2f%n", result);
        }
        
        sc.close();
    }
}
import math

class Solution:
    def calculate_sum(self, n: int, m: int) -> float:
        sum_val = 0
        current = n
        
        # 计算前m项的和
        for _ in range(m):
            sum_val += current
            current = math.sqrt(current)
        
        return sum_val

def main():
    solution = Solution()
    
    # 处理多组输入
    while True:
        try:
            n, m = map(int, input().split())
            result = solution.calculate_sum(n, m)
            # 格式化输出
            print(f"{result:.2f}")
        except EOFError:
            break

if __name__ == "__main__":
    main()

算法及复杂度

  • 算法:直接模拟
  • 时间复杂度:,其中 是数列项数
  • 空间复杂度:,只需要常数空间