子数组的解释与专题

//暴力
class Solution {
public:
    int countCompleteSubarrays(vector<int>& nums) 
    {
        int sum=0;
        set<int> s;
        for(auto& x:nums)
        s.insert(x);
        int l=nums.size();
        for(int i=0;i<l;i++)
        {
            unordered_set<int> ss;
            for(int j=i;j<l;j++)
            {
                ss.insert(nums[j]);
                if(s.size()==ss.size())
                sum++;
            }
        }
        return sum;
    }
};

//滑动窗口
class Solution {
public:
    int countCompleteSubarrays(vector<int> &nums) {
        int m = unordered_set<int>(nums.begin(), nums.end()).size();
        unordered_map<int, int> cnt;
        int ans = 0, left = 0;
        for (int v: nums) { // 枚举子数组右端点 v=nums[i]
            cnt[v]++;
            while (cnt.size() == m) {
                int x = nums[left++];
                if (--cnt[x] == 0)
                    cnt.erase(x);
            }
            ans += left; // 子数组左端点 < left 的都是合法的
        }
        return ans;
    }
};

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5+10;
int a[N],n,p,sum[N],ans;
int sumn;
void solve()
{
    cin>>n>>p;
    for(int i=0;i<n;i++){
        cin>>a[i];
        sumn += a[i];
    }
    sum[0] = a[0];
    for(int i=1;i<n;i++){
        sum[i] = sum[i-1]+a[i];
    }
    if(n == 2){
        int cnt = a[0] % p + a[1] % p; 
        cout<<cnt<<endl;
        return ;
    }
    for(int i=1;i<n-1;i++){
        int tmp = sum[i-1]%p+(sumn-sum[i-1])%p;
        ans = max(ans,tmp);
    }
    cout<<ans<<endl;
    return ;
}
signed main()
{
  int t=1;
  while(t--)
  solve();
  return 0;
}