我的题解

这是我的正确代码：

#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m;
	cin >> n >> m;
	vector<int> number(n), color(n);
	for (int i = 0; i < n; i++) cin >> number[i];
	for (int i = 0; i < n; i++) cin >> color[i];
	unordered_map<int, vector<int>> groups;
	for (int i = 0; i < n; i++) {
		groups[color[i]].push_back(i);
	}
	long long ans = 0;
	const int MOD = 10007;
	for (auto &p : groups) {
		auto &indices = p.second;
		for (int i = 0; i < indices.size(); i++) {
			for (int j = i + 1; j < indices.size(); j++) {
				int x = indices[i], z = indices[j];
				if ((x + z) % 2 == 0) {
					long long id_sum = (x + 1) + (z + 1);
					long long num_sum = number[x] + number[z];
					ans = (ans + id_sum * num_sum) % MOD;
				}
			}
		}
	}
	cout << ans % MOD << endl;
	return 0;
}

反正想的是分组算，这样再到更小的组里枚举时间复杂度就不会炸了。

标准的错误答案（我之前写的）：

#include<bits/stdc++.h>
int main() {
	int x, y, z;
	int n, m;
	std::cin >> n >> m;
	int number[n], color[n];
	for (int i = 0; i < n; i++) {
		std::cin >> number[i];
	}
	for (int i = 0; i < n; i++) {
		std::cin >> color[i];
	}
	std::vector<std::pair<int, int>> pairsForColors;
	for (int x = 0; x < n; x++) {
		for (int z = x + 1; z < n; z++) {
			if (color[x] == color[z]) {
				pairsForColors.push_back({x + 1, z + 1});
			}
		}
	}
	std::set<std::pair<int, int>> founded;
	for (int x = 0; x < pairsForColors.size(); x++) {
		for (int y = 2; y <= 2 * n; y += 2) {
			if ((pairsForColors[x].first + pairsForColors[x].second) == y) {
				founded.insert({pairsForColors[x].first, pairsForColors[x].second});
			}
		}
	}
	int itTurnsOut = 0;
	for (std::pair<int, int> x : founded) {
		itTurnsOut += ((x.first + x.second) * (number[x.first - 1] + number[x.second - 1]));
	}
	std::cout << itTurnsOut % 10007;
}

之前写的还会MLE、WA。反正看得懂的都看懂了