Overview

This problem asks us to calculate the sum of forecast errors \(S_i = \sum_{j=1}^{D} |A_{i,j} - T_i|\) for each location \(i\) and find the maximum value among them. You can solve it by straightforwardly calculating according to the definition in the problem statement.

Analysis

Since the forecast temperatures \(T_i\) are already given, there is no need to think about optimization or clever tricks. What we need to do is simply calculate the sum of forecast errors for each location according to the definition formula and find the maximum value.

The important point to consider here is computational complexity and input size.

• The number of locations \(N\) and the number of days \(D\) are at most \(10000\) each, but there is a constraint of \(N \times D \leq 10^6\).
• This means that the total number of operations to read and process all temperature data \(A_{i,j}\) one by one is only about \(10^6\).

Therefore, even if we scan all the data honestly, it can be processed fast enough, so no special algorithm is required.

On the other hand, we should be careful about integer overflow. Since the temperature is \(-50 \leq A_{i,j}, T_i \leq 50\), the error per day is at most \(100\). If this accumulates over \(D = 10000\) days, \(S_i\) can be up to \(100 \times 10000 = 10^6\). While this value fits in a standard 32-bit signed integer (int), using long long (a 64-bit integer) is safe and recommended.

Algorithm

The steps are as follows:

1. Read \(N\) (the number of locations) and \(D\) (the number of days).
2. Read the forecast temperatures \(T_1, \ldots, T_N\) for each location.
3. For each location \(i\):
   • Initialize the total error \(s\) to \(0\).
   • Read the actual temperatures \(A_{i,j}\) for \(D\) days in order, and add \(|A_{i,j} - T_i|\) to \(s\).
   • If the calculated \(s\) is greater than the maximum value \(ans\) found so far, update \(ans\).
4. Finally, output \(ans\).

The key point is that we do not need to store all the data for each location in memory. Since we can calculate the sum sequentially while reading the input, we can process it without maintaining additional arrays.

Concrete Example: If the forecast temperature for location \(i\) is \(T_i = 20\) and the actual temperatures are \([22, 18, 25]\), then $\(S_i = |22-20| + |18-20| + |25-20| = 2 + 2 + 5 = 9\)$ We calculate this for all locations and output the maximum.

Complexity

• Time Complexity: \(O(N \times D)\) (at most around \(10^6\) operations due to the constraints)
• Space Complexity: \(O(N)\) (only for the forecast temperature array; actual temperature data can be discarded immediately after reading, so no array is needed for it)

Implementation Points

• Type of the sum: Since the maximum value of \(S_i\) can reach around \(10^6\), using long long is safe and reassuring.

• Absolute value: Calculate the forecast error using abs(a - t).

• Fast I/O: Since we read \(10^6\) integers, it might take some time if the input is slow. In C++, using scanf or setting ios::sync_with_stdio(false) when using cin is recommended for fast I/O.

• Memory saving: The temperature data \(A_{i,j}\) can be used for calculation immediately after reading, so there is no need to store it in an array. This keeps the memory usage low.

  Source Code

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

int main(){
    int N, D;
    scanf("%d %d", &N, &D);
    vector<int> T(N);
    for(int i=0;i<N;i++) scanf("%d", &T[i]);
    long long ans = 0;
    for(int i=0;i<N;i++){
        long long s = 0;
        int t = T[i];
        for(int j=0;j<D;j++){
            int a;
            scanf("%d", &a);
            s += abs(a - t);
        }
        ans = max(ans, s);
    }
    printf("%lld\n", ans);
    return 0;
}

This editorial was generated by claude4.8opus-high.