题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6621
题意为求区间[l,r]内第k小|a[i]-p|的值。
可以二分答案，如果二分的值为x，则判断区间[l,r]内是否有k个数在[p-x,p+x]范围内。所以就用主席树搞一下。
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<string> 5 #include<cmath> 6 #include<vector> 7 #define lson l, mid, i<<1 8 #define rson mid + 1, r, i<<1|1 9 using namespace std; 10 typedef long long ll; 11 const int maxn = 1e6 + 10; 12 int root[maxn], rs[maxn * 20], ls[maxn * 20], val[maxn * 20]; 13 int cnt; 14 void update(int k, int l, int r, int &i) { 15 val[++cnt] = val[i] + 1, ls[cnt] = ls[i], rs[cnt] = rs[i]; 16 i = cnt; 17 if (l == r) 18 return; 19 int mid = l + r >> 1; 20 if (k <= mid) 21 update(k, l, mid, ls[i]); 22 else 23 update(k, mid + 1, r, rs[i]); 24 } 25 int query(int u, int v, int L, int R, int l, int r) { 26 if (L <= l && r <= R) 27 return val[u] - val[v]; 28 int mid = l + r >> 1; 29 int ans = 0; 30 if (L <= mid) 31 ans += query(ls[u], ls[v], L, R, l, mid); 32 if (R > mid) 33 ans += query(rs[u], rs[v], L, R, mid + 1, r); 34 return ans; 35 } 36 int check(int L, int R, int p, int x) { 37 int l = max(p - x, 1), r = min(p + x, (int)1e6); 38 return query(root[R], root[L - 1], l, r, 1, 1e6); 39 } 40 int main() { 41 int t; 42 scanf("%d", &t); 43 while (t--) { 44 int n, m, x, cnt = 0; 45 scanf("%d%d", &n, &m); 46 for (int i = 1; i <= n; i++) { 47 scanf("%d", &x); 48 root[i] = root[i - 1]; 49 update(x, 1, 1e6, root[i]); 50 } 51 int ans = 0, L, R, p, k; 52 for (int i = 1; i <= m; i++) { 53 scanf("%d%d%d%d", &L, &R, &p, &k); 54 L ^= ans, R ^= ans, p ^= ans, k ^= ans; 55 int l = 0, r = 1e6; 56 while (l <= r) { 57 int mid = l + r >> 1; 58 int w = check(L, R, p, mid); 59 if (w >= k) { 60 r = mid - 1; 61 ans = mid; 62 } 63 else 64 l = mid + 1; 65 } 66 printf("%d ", ans); 67 } 68 } 69 }