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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#876676#7627. PhonychangziliangWA 0ms8016kbC++145.7kb2025-01-31 10:59:412025-01-31 10:59:52

Judging History

This is the latest submission verdict.

  • [2025-01-31 10:59:52]
  • Judged
  • Verdict: WA
  • Time: 0ms
  • Memory: 8016kb
  • [2025-01-31 10:59:41]
  • Submitted

answer

#include<bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
typedef long long LL;
typedef __int128 Int;
int n, m;
LL K, a[N], Lim;
namespace Segment {
	struct SegmentTree {
		int ls, rs, cnt;
		#define ls(x) tree[x].ls
		#define rs(x) tree[x].rs
		#define cnt(x) tree[x].cnt
	}tree[N * 40];
	int tot;
	int newnode() {return ++ tot;}
	void update(int p) {
		cnt(p) = cnt(ls(p)) + cnt(rs(p));
	}
	void ins(int &p, LL lp, LL rp, LL pos) {
		if(!p) p = newnode();
		if(lp == rp) {
			cnt(p) ++;
			return ;
		}
		LL mid = (lp + rp >> 1);
		if(pos <= mid) ins(ls(p), lp, mid, pos);
		else ins(rs(p), mid + 1, rp, pos);
		update(p);
	}
	LL ask(int p, LL lp, LL rp, int x) {
		//cout << "GGG" << ' ' << lp << ' ' << rp
		if(lp == rp) return lp;
		LL mid = (lp + rp >> 1);
		if(cnt(ls(p)) >= x) return ask(ls(p), lp, mid, x);
		else return ask(rs(p), mid + 1, rp, x - cnt(ls(p)));
 	}
 	int Merge(int p, int q, LL lp, LL rp) {
 		if(!p || !q) return p ^ q;
 		if(lp == rp) {
 			cnt(p) += cnt(q);
 			return p;
		}
		LL mid = (lp + rp >> 1);
		ls(p) = Merge(ls(p), ls(q), lp, mid);
		rs(p) = Merge(rs(p), rs(q), mid + 1, rp);
		update(p);
		return p;
	}
	void split(int &p, int &q, LL lp, LL rp, LL c) {
		if(!p) return ;
		if(lp == rp) {
			if(cnt(p) == c) {
				q = p; p = 0;
				return ;
			}
			else {
				int u = newnode();
				cnt(u) = c; cnt(p) -= c;
			    q = u;
				return ;
			}
		}
		if(!q) q = newnode();
		LL mid = (lp + rp >> 1);
		if(cnt(rs(p)) >= c) split(rs(p), rs(q), mid + 1, rp, c);
		else {
			rs(q) = rs(p); rs(p) = 0;
			split(ls(p), ls(q), lp, mid, c - cnt(rs(q)));
		}
		update(p); update(q);
	}
}
namespace Fhq { // 改成平衡树   空间线性 
	int root, tot;
	int rt1, rt2, rt3;
	mt19937_64 rnd(time(0));
	struct Node {
		int son[2];
		int cnt, sz, rt; // 数量, 根 
		LL val, w; // 数值  随机权重 
	}t[N * 2];
	int newnode(LL v) {
		tot ++;
		t[tot].val = v;
		t[tot].w = rnd();
		return tot;
	}
	void ins(int p, LL v) {
		Segment::ins(t[p].rt, 0, K - 1, v);
		t[p].cnt ++; t[p].sz ++;
	}
	void update(int p) {
		t[p].sz = t[t[p].son[0]].sz + t[t[p].son[1]].sz + t[p].cnt;
	}
	int Merge(int p, int q) {
	    if(!p || !q) return p ^ q;	
	    if(t[p].w > t[q].w) {
	    	t[p].son[1] = Merge(t[p].son[1], q);
	    	update(p);
	    	return p;
		}
		else {
			t[q].son[0] = Merge(p, t[q].son[0]);
			update(q);
			return q;
		}
	}
	void split(int p, int &x, int &y, LL lim) {
		if(!p) {x = y = 0; return ;}
		if(t[p].val <= lim) x = p, split(t[p].son[1], t[p].son[1], y, lim);
		else y = p, split(t[p].son[0], x, t[p].son[0], lim);
		update(p);
	}
	void add(LL v1, LL v2) { // 插入一个 
		split(root, rt1, rt2, v1);
		split(rt1, rt1, rt3, v1 - 1);
		if(!rt3) rt3 = newnode(v1);
		ins(rt3, v2);
		root = Merge(Merge(rt1, rt3), rt2);
	}
	LL query(int p, int x) { // 查询第 x 个 
	  //  if(!p) return 0;
		if(t[t[p].son[0]].sz >= x) return query(t[p].son[0], x);
		else if(t[t[p].son[0]].sz + t[p].cnt >= x) return 1LL * t[p].val * K + Segment::ask(t[p].rt, 0, K - 1, x - t[t[p].son[0]].sz);
		else return query(t[p].son[1], x - (t[t[p].son[0]].sz + t[p].cnt));
	}
	LL Find(int p, LL x) {
		if(!p) return -2e18;
		if(t[p].son[1] && t[t[p].son[1]].val <= x) return Find(t[p].son[1], x);
		else if(t[p].val <= x) return t[p].val;
		else return Find(t[p].son[0], x);
 	}
 	LL Count(int p, LL x) {
 		if(t[p].val == x) return t[p].cnt;
 		if(t[p].val > x) return Count(t[p].son[0], x);
 		else return Count(t[p].son[1], x);
	}
	int del(LL x) { // 删去值为 x 的节点,并且返回它的根 
		split(root, rt1, rt2, x);
		split(rt1, rt1, rt3, x - 1); // 删掉 rt3 的根 
		root = Merge(rt1, rt2);
		return t[rt3].rt;
	}
	void Crt(LL x, int p) {
		split(root, rt1, rt2, x);
		split(rt1, rt1, rt3, x - 1);
		if(!rt3) rt3 = newnode(x);
		if(t[rt3].sz > 0) {
			puts("HHH");
		}
		t[rt3].rt = p; 
		t[rt3].sz = t[rt3].cnt = Segment::cnt(p);
		root = Merge(Merge(rt1, rt3), rt2);
	}
	void Mg(LL x, LL y) { // 把 y 删掉, 合并到 x 里面 
		int p = del(x), q = del(y);
		p = Segment::Merge(p, q, 0, K - 1);
		Crt(x, p);
	}
	void Sp(LL x, LL c) {
		int p = del(x), q = del(x - 1);
		int Rt = 0;
		Segment::split(p, Rt, 0, K - 1, c);
		Crt(x, p); q = Segment::Merge(q, Rt, 0, K - 1);
		Crt(x - 1, q);
	}
}
void move(LL x) { // 将最大的减少K  x次    方法是每次找到最后一个1,看能不能和前面那个合并,然后在分裂一次 
	while(x) {
		LL lst = Fhq::Find(Fhq::root, 1e18); // 找到 
		LL pre = Fhq::Find(Fhq::root, lst - 1);
		LL ct = Fhq::Count(Fhq::root, lst);
		if(pre >= -1e18 && (Int)(lst - pre) * ct <= x) {
			x -= (lst - pre) * ct;
			Fhq::Mg(pre, lst);
			continue;
		}
		if(Fhq::t[Fhq::root].sz != n) {
			printf("MG\n");
			exit(0);
		}
		LL step = x / ct; // 整体移动 step 
		int rt = Fhq::del(lst);
		Fhq::Crt(lst - step, rt); // 改变根 
		x -= ct * step; // 还剩下的需要分裂 
		if(Fhq::t[Fhq::root].sz != n) {
			printf("move\n");
			exit(0);
		}
		if(x) Fhq::Sp(lst - step, x); // 分裂
		if(Fhq::t[Fhq::root].sz != n) {
			printf("split\n");
			exit(0);
		}
		x = 0;
	}
}
int main() {
	scanf("%d%d%lld", &n, &m, &K);
	for(int i = 1; i <= n; i ++ ) {
		scanf("%lld", &a[i]);
		Fhq::add(a[i] / K, a[i] % K);
	}
	for(int i = 1; i <= m; i ++ ) {
		char opt; LL x;
		scanf("\n%c%lld", &opt, &x);
		if(opt == 'C') { // 修改 
			move(x); // 移动 x 步 
		} 
		else { // 查询第x大 
			if(Fhq::t[Fhq::root].sz == n) printf("%lld\n", Fhq::query(Fhq::root, n - x + 1));
			else {
				printf("CNM %d\n", Fhq::t[Fhq::root].sz);
				exit(0);
			}
		}
	}
	return 0;
}
/*
3 5 5
7 3 9
A 3
C 1
A 2
C 2
A 3
*/

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 8016kb

input:

3 5 5
7 3 9
A 3
C 1
A 2
C 2
A 3

output:

3
4
-1

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 8016kb

input:

5 8 8
294 928 293 392 719
A 4
C 200
A 5
C 10
A 2
C 120
A 1
A 3

output:

294
200
191
0
-2

result:

ok 5 lines

Test #3:

score: -100
Wrong Answer
time: 0ms
memory: 8016kb

input:

100 100 233
5101 8001 6561 6329 6305 7745 4321 811 49 1121 3953 8054 8415 9876 6701 4097 6817 6081 495 5521 2389 2042 4721 8119 7441 7840 8001 5756 5561 129 1 5981 4801 7201 8465 7251 6945 5201 5626 3361 5741 3650 7901 2513 8637 3841 5621 9377 101 3661 5105 4241 5137 7501 5561 3581 4901 561 8721 811...

output:

HHH
move

result:

wrong answer 1st lines differ - expected: '6881', found: 'HHH'