A - Distributing Requests to Servers
【题目来源】
AtCoder：A - Distributing Requests to Servers
【题目描述】
Takahashi is a web service operations manager. His system has \(N\) servers, and a large-scale campaign spanning \(M\) days starts tomorrow.
高桥是一名网络服务运维经理。他的系统有 \(N\) 台服务器，一场持续 \(M\) 天的大规模活动将于明天开始。
During the campaign period, exactly \(1\) heavy processing request occurs each day. For each of the \(M\) days, Takahashi must assign that request to exactly \(1\) of the \(N\) servers. The same server may be chosen on multiple days.
在活动期间，每天恰好发生 \(1\) 个重处理请求。对于这 \(M\) 天中的每一天，高桥必须将该请求分配给 \(N\) 台服务器中的恰好 \(1\) 台。同一台服务器可在多天被选择。
Each server \(i\) (\(1 \leq i \leq N\)) has a "processing capacity" \(A_i\). If the total number of requests assigned to server \(i\) throughout the entire campaign period is at most \(A_i\), that server operates normally. On the other hand, if more than \(A_i\) requests are assigned to it, that server will go down. Takahashi must not let any server go down.
每台服务器 \(i\)（\(1 ≤ i ≤ N\)）具有一个"处理容量" \(A_i\)。如果分配给服务器 \(i\) 的请求总数在整个活动期间不超过 \(A_i\)，则该服务器正常运行。反之，如果分配给它的请求超过 \(A_i\)，则该服务器将宕机。高桥绝不能允许任何服务器宕机。
Can Takahashi choose request assignments wisely enough to get through the \(M\)-day campaign without any server going down?
高桥能否明智地选择请求分配方案，以便在 \(M\) 天的活动期间不让任何服务器宕机？
If there exists an assignment that allows all \(M\) days to be completed without any server going down, output Yes; otherwise, output No.
如果存在一种分配方案使得所有 \(M\) 天都能完成且不让任何服务器宕机，则输出 Yes；否则，输出 No。
【输入】
\(N\) \(M\)
\(A_1\) \(A_2\) \(\ldots\) \(A_N\)
- The first line contains an integer \(N\) representing the number of servers and an integer \(M\) representing the number of campaign days, separated by a space.
- The second line contains \(N\) integers \(A_1, A_2, \ldots, A_N\) representing the processing capacity of each server, separated by spaces.
【输出】
If there exists an assignment that allows all \(M\) days of the campaign to be completed without any server going down, output Yes; otherwise, output No on a single line.
【输入样例】
3 10
3 4 2
【输出样例】
No
【解题思路】
【代码详解】
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n, m, sum; // n: 数字个数，m: 目标值，sum: 数字总和signed main()
{cin >> n >> m; // 读入数字个数n和目标值mfor (int i = 1; i <= n; i++) // 读取n个数字{int x; cin >> x; // 读入第i个数字sum += x; // 累加到总和}if (sum >= m) // 如果总和大于等于目标值{cout << "Yes";}else{cout << "No";}return 0;
}
【运行结果】
3 10
3 4 2
No