#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> calculate_costs(vector<int> W, vector<int> A, vector<int> B, vector<int> E) {
int n = W.size();
int q = E.size();
vector<ll> R(q, 0);
ll sb = 0;
// Calculate sum of B
for (int i = 0; i < n; i++) {
sb += B[i];
}
for (int i = 0; i < q; i++) {
ll d = E[i];
long long pmin = LLONG_MAX;
// Check if third subtask conditions are met
bool third_subtask = (q <= 5 && all_of(A.begin(), A.end(), [](int a) { return a == 2; }) && all_of(B.begin(), B.end(), [](int b) { return b == 1; }));
if (third_subtask) {
// If third subtask, apply optimized calculation
R[i] = sb + n; // Since A[i] - B[i] = 1 for all i, and we need to add sum(B) + min(A[i] - B[i])
continue;
}
if (n % 2 == 0) {
// If n is even, set result to sum of B values
R[i] = sb;
continue;
} else {
// Otherwise, find minimum (A[j] - B[j]) under the given conditions
for (int j = 0; j < n; j++) {
if (j % 2 == 0) {
pmin = min(pmin, (long long)A[j] - B[j]);
}
if (j > 0 && j < n - 1) {
if (W[j + 1] - W[j - 1] <= d) {
pmin = min(pmin, (long long)A[j] - B[j]);
}
}
}
}
R[i] = pmin + sb;
}
return R;
}
int main() {
int n, q;
cin >> n >> q;
vector<int> W(n), A(n, 2), B(n, 1), E(q);
// Input values for W, A, B arrays
for (int i = 0; i < n; i++) {
W[i] = i + 1;
cin >> A[i] >> B[i];
}
// Input values for E array
for (int i = 0; i < q; i++) {
cin >> E[i];
}
// Get result for each query
vector<ll> result = calculate_costs(W, A, B, E);
// Print the results
for (size_t i = 0; i < result.size(); i++) {
cout << result[i] << " ";
}
cout << endl;
return 0;
}
Compilation message
/usr/bin/ld: /tmp/ccsWq3XU.o: in function `main':
grader.cpp:(.text.startup+0x0): multiple definition of `main'; /tmp/ccYb8TLR.o:nile.cpp:(.text.startup+0x0): first defined here
collect2: error: ld returned 1 exit status
