이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <map>
#include <unordered_map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#include <iomanip>
using namespace std;
#define ll long long
#define PLL pair<ll, ll>
#define PB push_back
#define F first
#define S second
#define MP make_pair
const ll INF = 999999999999999999;
const ll M = 1000000000+7;
ll binExp(ll a, ll b){
if(a==1){
return 1;
}
if(b==0){
return 1;
}
ll ans = 1;
if(b%2==1){
ans = a;
}
return (binExp((a*a)%M,b/2)*ans)%M;
}
// t.F -> grab indexes 0,2,4,...
// t.S -> grab indexes 1,3,5,...
void build(ll v, ll l, ll r, vector<PLL> &t, vector<ll> &a){
if(l==r){
t[v].F = a[l];
t[v].S = 0;
return;
}
ll m = (l+r)/2;
build(2*v,l,m,t,a);
build(2*v+1,m+1,r,t,a);
if(l%2!=m%2){
t[v].F = t[2*v].F^t[2*v+1].F;
t[v].S = t[2*v].S^t[2*v+1].S;
}
else{
t[v].F = t[2*v].F^t[2*v+1].S;
t[v].S = t[2*v].S^t[2*v+1].F;
}
}
void update(ll v, ll l, ll r, ll lx, ll pr, ll x, vector<PLL> &t){
if(l>lx or r<lx){
return;
}
if(l%2==lx%2){
t[v].F = t[v].F^pr^x;
}
else{
t[v].S = t[v].S^pr^x;
}
if(l==r){
return;
}
ll m = (l+r)/2;
update(2*v,l,m,lx,pr,x,t);
update(2*v+1,m+1,r,lx,pr,x,t);
}
ll XoR(ll v, ll l, ll r, ll lx, ll rx, vector<PLL> &t){
//cout<<v<<" "<<l<<" "<<r<<endl;
if(lx<=l and r<=rx){
if((lx%2)==(l%2)){
return t[v].F;
}
else{
return t[v].S;
}
}
if(l>rx or r<lx){
return 0;
}
ll m = (l+r)/2;
return (XoR(2*v,l,m,lx,rx,t)^XoR(2*v+1,m+1,r,lx,rx,t));
}
void solve(ll n, ll q, vector<ll> &a, vector<ll> &type, vector<PLL> &values){
vector<PLL> t(4*n);
build(1,0,n-1,t,a);
for(ll i=0; i<q; i++){
//cout<<i<<endl;
if(type[i]==1){
values[i].F--;
update(1,0,n-1,values[i].F,a[values[i].F],values[i].S,t);
a[values[i].F] = values[i].S;
}
else{
values[i].F--;
values[i].S--;
if((values[i].S-values[i].F+1)%2==0){
cout<<0<<endl;
}
else{
cout<<XoR(1,0,n-1,values[i].F,values[i].S,t)<<endl;
}
}
}
}
int main(){
ll n,q;
cin>>n>>q;
vector<ll> a(n);
for(ll i=0; i<n; i++){
cin>>a[i];
}
vector<ll> type(q);
vector<PLL> ranges(q);
for(ll i=0; i<q; i++){
cin>>type[i]>>ranges[i].F>>ranges[i].S;
}
solve(n,q,a,type,ranges);
}
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