做法 1

$$\begin{gathered} (x+z)\times (nu{m}_x+nu{m}_z) \\ =x\times nu{m}_x+x\times nu{m}_z+z\times nu{m}_x+z\times nu{m}_z \end{gathered}$$

#include <bits/stdc++.h>
using namespace std;
const int MODNUM = 10007;
int n, m;
int number[100005];
int color[100005];
int sumOdd[100005];
vector<int> odd[100005];
int sumEven[100005];
vector<int> even[100005];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> number[i];
        number[i] %= MODNUM;
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> color[i];
        if (i % 2)
            odd[color[i]].push_back(i), 
            sumOdd[color[i]] = (sumOdd[color[i]] + number[i]) % MODNUM;
        else
            even[color[i]].push_back(i), 
            sumEven[color[i]] = (sumEven[color[i]] + number[i]) % MODNUM;
    }
    int ans = 0;
    for (int i = 1; i <= m; i++)
    {
        for (int j = 0; j < odd[i].size(); j++)
        {
            int x = odd[i][j];
            ans = (ans + x * (sumOdd[i] - number[x] + MODNUM) % MODNUM) % MODNUM;
            ans = (ans + (odd[i].size() - 1) * x % MODNUM * number[x] % MODNUM) % MODNUM;
        }
        for (int j = 0; j < even[i].size(); j++)
        {
            int x = even[i][j];
            ans = (ans + x * (sumEven[i] - number[x] + MODNUM) % MODNUM) % MODNUM;
            ans = (ans + (even[i].size() - 1) * x % MODNUM * number[x] % MODNUM) % MODNUM;
        }
    }
    cout << ans << "\n";
    return 0;
}