把询问离线, 从小到大解决, 转换成求第k大的问题, 套个平衡树就好啦。
#include<bits/stdc++.h> #include <bits/extc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; using namespace __gnu_pbds; const int N = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} template <class T> using Tree = tree<T, null_type, std::less<T>, rb_tree_tag,tree_order_statistics_node_update>; Tree<int> bst; int n, m, q; PLL qus[N]; PLL now[N]; int a[N]; int cnt[N]; int ans[N]; int main() { scanf("%d%d%d", &n, &m, &q); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); now[a[i]].fi++; } for(int i = 1; i <= m; i++) now[i].se = i; sort(now + 1, now + 1 + m); for(int i = 1; i <= q; i++) { scanf("%lld", &qus[i].fi); qus[i].se = i; } sort(qus + 1, qus + 1 + q); LL pre = 0; LL cnt = 0; LL tmp = n; for(int i = 1, j = 1; i <= q; i++) { LL k = qus[i].fi; while(j <= m && (now[j].fi - pre) * cnt + tmp < k) { tmp += (now[j].fi - pre) * cnt; cnt++; bst.insert(now[j].se); pre = now[j].fi; j++; } LL need = k - tmp; LL pos = need % cnt; if(pos) pos = *bst.find_by_order(pos - 1); else pos = *bst.find_by_order(cnt - 1); ans[qus[i].se] = pos; } for(int i = 1; i <= q; i++) printf("%d\n", ans[i]); return 0; } /* */