题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5, base = 397;
const int mod = 1e9 + 7;
// const int mod = 998244353;
// const __int128 mod = 212370440130137957LL;
const int N = 1e5;
// int a[1005][1005];
int a[maxn], b[maxn];
// bool vis[maxn];
int n, m;
string s;struct Node{// int val, id;// bool operator<(const Node &u)const{//     return val < u.val;// }int a, b;bool operator<(const Node &u)const{return b < u.b;}
}c[maxn];void solve(){int res = 0;int k, q;int c;cin >> n >> c;for(int i = 1; i <= n; i++){cin >> a[i];}int odd = 0, even = 0;res = (c + 1) * c / 2 + (c + 1);//x，y可以相同int sum = 0;for(int i = 1; i <= n; i++){sum += a[i] / 2 + 1;sum += c - a[i] + 1;if(a[i] % 2 == 1){odd++;}else{even++;}}// cout << sum << '\n';sum -= even * (even + 1) / 2 + odd * (odd + 1) / 2;//可以选两个相同的（a[i] <= c, 所以        //y = (a[i] + a[j]) / 2必然小于等于cres -= sum;cout << res << '\n';
}   signed main(){ios::sync_with_stdio(0);cin.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}