Overview

Over a period of \(M\) days, a total of \(M\) requests occur one per day, and we must assign them to \(N\) servers. Each server \(i\) has a maximum number of requests it can handle, \(A_i\). The problem is to determine whether all requests can be assigned without exceeding any server’s limit.

Analysis

The key insight for solving this problem is to consider “what is the maximum total number of requests the entire system can handle?”

Given that the maximum number of requests each server can handle is \(A_1, A_2, \ldots, A_N\), the total number of requests the entire system can accept is the sum of all these values.

• When all requests can be handled If the total processing capacity of the entire system is at least \(M\) (the total number of requests that occur during the campaign period), then by appropriately distributing requests to each server, we can get through without any server going down. (For example: each time a request arrives, simply choose any server that still has remaining capacity and assign it there.)

• When requests cannot all be handled If the total processing capacity of the servers is less than \(M\), then no matter how requests are distributed, eventually some server will exceed its limit, or a request will be left unprocessable.

Therefore, this problem can be reduced to simply determining “whether the sum of \(A_i\) is at least \(M\).”

Algorithm

1. Read the input values \(N, M\) and the processing capacity \(A_i\) for each server.
2. Compute the total sum \(S = \sum_{i=1}^N A_i\).
3. If \(S \geq M\), output Yes; otherwise, output No.

Complexity

• Time complexity: \(O(N)\)
  • We read \(N\) numerical values for the servers and compute their sum, which takes time proportional to \(N\).
• Space complexity: \(O(N)\)
  • If the input values \(A_i\) are stored in a list or similar structure, memory proportional to \(N\) is used.

Implementation Notes

• Handling large numbers: The values of \(M\) and \(A_i\) can be as large as \(10^9\), which is very large. However, in Python 3, integers have no size limit, so we can compute the sum and compare directly without worrying about overflow.

• Speeding up input: Since \(N\) can be as large as \(10^5\), we need to read a large number of values. In Python, reading all input at once using sys.stdin.read().split() is faster than calling input() repeatedly.

  Source Code

import sys

def solve():
    # 標準入力からNとMを読み込む
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    n = int(input_data[0])
    m = int(input_data[1])
    
    # サーバーの処理可能回数A_iを読み込む
    a = list(map(int, input_data[2:]))
    
    # 全サーバーの処理可能回数の合計を計算する
    total_capacity = sum(a)
    
    # 合計がキャンペーン期間中のリクエスト数M以上であれば、割り当てが可能
    if total_capacity >= m:
        print("Yes")
    else:
        print("No")

if __name__ == "__main__":
    solve()

This editorial was generated by gemini-3-flash-thinking.