题意：给一个序列a，以及K，有Q个询问，每个询问四个数，L,R,U,V, 求L<=i<=R,U<=j<=V，a[i]+a[j]=K的(i, j)对数（题目保证了L <= R < U <= V）。
思路：首先用容斥原理把询问变为i,j同区间，记f(A,B)为答案，'+'为区间的并，A=[L,R],B=[U,V],C=[u+1,v-1]，则f(A,B) = f(A+B+C,A+B+C)+f(C,C)-f(A+C,A+C)-f(B+C,B+C)。令g(L,R) = f([L,R],[L,R]) = f(A,A)。对于g函数区间变化为1时，可以做到o(1)的维护。用莫队算法的知识把询问排个序，按顺序维护当前区间的答案，同时更新最后答案。
总结：莫队算法并不是某一具体问题的解法，而是一种通用算法，对于任意无修改的区间统计问题，都可以用莫队算法试试。对于所有询问(L,R)，复杂度等于sigma(|Li - Li-1| + |Ri - Ri-1|)*o(x)，o(x)是区间变化为1时的维护代价，而前面的sigma用莫队算法的知识可以降低到o(m*sqrt(n))。
#pragma comment(linker, "/STACK:10240000,10240000") #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility> using namespace std; #define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d\n", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a) typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi; const int dx[] = {, , -, , , , -, -};
const int dy[] = {-, , , , , -, , - };
const int maxn = 3e4 + ;
const int md = ;
const int inf = 1e9 + ;
const LL inf_L = 1e18 + ;
const double pi = acos(-1.0);
const double eps = 1e-; template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; } struct Node {
int L, R, id, kth;
constructInt4(Node, L, R, id, kth);
};
Node node[maxn * ]; int block;
int a[maxn], ans[maxn], cnt[maxn]; bool cmp(const Node a, const Node b) {
int lb = a.L / block, rb = b.L / block;
return lb < rb || lb == rb && a.R < b.R;
} int main() {
//freopen("in.txt", "r", stdin);
int n, m, k;
while (cin >> n) {
cin >> k;
rep_up0(i, n) {
sint(a[i]);
}
cin >> m;
rep_up0(i, m) {
int L, R, U, V;
scanf("%d %d %d %d", &L, &R, &U, &V);
L --; R --; U --; V --;
node[i * ] = Node(L, V, i, );
node[i * + ] = Node(R + , U - , i, );
node[i * + ] = Node(L, U - , i, );
node[i * + ] = Node(R + , V, i, );
} block = (int)sqrt(n + 0.5);
sort(node, node + * m, cmp); int cur_ans = , L = , R = -;
mem0(cnt);
mem0(ans); rep_up0(i, * m) {
Node query = node[i];
while (L < query.L) {
cnt[a[L]] --;
if (k > a[L] && k - a[L] <= n) cur_ans -= cnt[k - a[L]];
L ++;
}
while (R > query.R) {
cnt[a[R]] --;
if (k > a[R] && k - a[R] <= n) cur_ans -= cnt[k - a[R]];
R --;
}
while (L > query.L) {
L --;
if (k > a[L] && k - a[L] <= n) cur_ans += cnt[k - a[L]];
cnt[a[L]] ++;
}
while (R < query.R) {
R ++;
if (k > a[R] && k - a[R] <= n) cur_ans += cnt[k - a[R]];
cnt[a[R]] ++;
}
if (query.kth < ) ans[query.id] += cur_ans;
else ans[query.id] -= cur_ans;
} rep_up0(i, m) {
printf("%d\n", ans[i]);
}
}
return ;
}