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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
#176437#6427. Just Another Game of StoneshazeRE 60ms153904kbC++143.1kb2023-09-11 17:10:252023-09-11 17:10:25

Judging History

你现在查看的是最新测评结果

  • [2023-09-11 17:10:25]
  • 评测
  • 测评结果:RE
  • 用时:60ms
  • 内存:153904kb
  • [2023-09-11 17:10:25]
  • 提交

answer

#include<bits/stdc++.h>
#define ll long long
#define irep(i,l,r) for(ll i = l; i <= r; ++i)
using namespace std;
inline ll read(){
	ll s=0; bool fl = 0;
	char chcc=getchar();
	while(chcc<'0'||chcc>'9'){if(chcc == '-')fl = 1;chcc = getchar();}
	while(chcc>='0'&&chcc<='9') s=(s<<3)+(s<<1)+(chcc^48),chcc=getchar();
	return fl?-s:s;
}
const int N = 800999;
const int itinf = 1000000000;
int n, m;
struct node{
	vector<int>bsum;
	int xorsum, mn, smn, len, tag;
	node(){bsum.resize(32);}
}t[N];
void fdmn(vector<int>aa, int &mn, int &smn){
	sort(aa.begin(), aa.end());
	unique(aa.begin(), aa.end());
	mn = aa[0], smn = aa[1];
}
void update(int x){
	fdmn({t[x*2].mn, t[x*2].smn, t[x*2+1].mn, t[x*2+1].smn}, t[x].mn, t[x].smn);
	t[x].len = 0, t[x].xorsum = t[x*2].xorsum ^ t[x*2+1].xorsum;
	if(t[x*2].mn == t[x].mn)t[x].len += t[x*2].len;
	if(t[x*2+1].mn == t[x].mn)t[x].len += t[x*2+1].len;
	irep(i,0,31)t[x].bsum[i] = t[x*2].bsum[i] + t[x*2+1].bsum[i];
}

void operate(int x,int lim){
	if(t[x].len & 1)t[x].xorsum ^= (lim ^ t[x].mn);
	t[x].tag = lim;
	int od = t[x].mn, nw = lim;
	irep(i,0,31){
		t[x].bsum[i] += ((nw & 1) - (od & 1)) * t[x].len;
		nw >>= 1, od >>= 1;
	}
	t[x].mn = lim;
}

void pushdown(int x){
	int tag = t[x].tag;
	t[x].tag = 0;
	if(t[x*2].mn < tag)operate(x*2,tag);
	if(t[x*2+1].mn < tag)operate(x*2+1,tag);
}

void build(int x,int l,int r){
	if(l == r){
		int val;
		val = t[x].mn = t[x].xorsum = read();
		t[x].len = 1;
		t[x].smn = t[x].tag = itinf;
		irep(i,0,31){
			t[x].bsum[i] += val & 1;
			val >>= 1;
		}
		return;
	}
	int mid = (l + r) >> 1;
	build(x * 2,l, mid);
	build(x * 2 + 1, mid + 1, r);
	update(x);
	return;
}
void modify(int x,int l,int r,int L,int R, int lim){
	// x -> max(x, val)
	if(l > R || L > r)return;
	if(L <= l && r <= R){
		if(lim <= t[x].mn)return;
		pushdown(x);
		if(lim < t[x].smn){
			operate(x,lim);
			return;
		}
		int mid = (l + r) >> 1;
		if(l == r)return;
		modify(x * 2,l, mid, L,R,lim);
		modify(x * 2 + 1, mid + 1, r ,L,R,lim);
		update(x);
		return;
	}
	pushdown(x);
	int mid = (l + r) >> 1;
	modify(x * 2,l, mid, L,R,lim);
	modify(x * 2 + 1, mid + 1, r ,L,R,lim);
	update(x);
	return;
}
int xor_query(int x,int l,int r,int L,int R){
	if(l > R || L > r)return 0;
	if(L <= l && r <= R)return t[x].xorsum;
	if(l != r)pushdown(x);
	int mid = (l + r) >> 1;
	return xor_query(x*2,l,mid,L,R) ^ xor_query(x*2+1,mid+1,r,L,R);
}
int bit_cnt(int x,int l, int r, int L ,int R,int id){
//	cerr << x << ' '<<l << ' ' << r<<' '<< id<<endl;
	if(l != r)pushdown(x);
	if(l > R || L > r)return 0;
	if(L <= l && r <= R)return t[x].bsum[id];
	int mid = (l + r) >> 1;
	return bit_cnt(x *2 ,l,mid,L,R,id) + bit_cnt(x*2+1,mid+1,r,L,R,id);
}
int main(){
	n = read(), m = read();
	build(1,1,n);
	while(m --){
		int op = read(), l = read(), r = read(), x = read();
		if(op == 1){
			modify(1,1,n,l,r,x);
		}else{
			ll key = xor_query(1,1,n,l,r);
			if((key ^ x) == 0){
				puts("0");
				continue;
			}
			int hb = log2(key ^ x);
			ll ans = bit_cnt(1,1,n,l,r,hb);
			if(x & (1 << hb))++ ans;
			printf("%lld\n",ans);
		}
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 60ms
memory: 153904kb

input:

5 4
1 2 1 4 1
2 1 3 1
1 2 4 3
2 2 4 4
2 1 4 4

output:

1
0
3

result:

ok 3 number(s): "1 0 3"

Test #2:

score: -100
Runtime Error

input:

200000 200000
962352030 173642520 1008864183 74920228 684681800 500911321 1001441054 257633652 185843534 59168654 317689197 731348417 123888883 708119712 340055368 876566011 980078202 969174443 814012870 715639041 596932238 173757742 314504576 1045746913 740811577 570187156 999816627 12441059 122507...

output:


result: