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QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#176445 | #6427. Just Another Game of Stones | haze | TL | 65ms | 229280kb | C++14 | 3.1kb | 2023-09-11 17:19:50 | 2023-09-11 17:19:50 |
Judging History
answer
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2,abm,mmx")
#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 = 1200999;
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){
if(l > R || L > r)return 0;
if(L <= l && r <= R)return t[x].bsum[id];
if(l != r)pushdown(x);
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: 65ms
memory: 229280kb
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
Time Limit Exceeded
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:
38889 57353 46659 19709 34617 92781 1211 755 16087 26119 10557 9165 7337 14363 9129 21117 93233 16733 64611 737 32715 25031 148859 17395 76263 56261 19825 93349 48429 22291 31645 9833 35195 95623 184389 39321 153 6067 6495 25369 9041 35785 39783 111237 151543 54289 121169 165785 101803 14803 15271 1...