| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 461145 | zaneyu | 말 (IOI15_horses) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*input
3
2 1 3
3 4 1
1
2 1 2
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=4e18;
const ll INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll sub(ll a,ll b){
ll x=a-b;
while(x<0) x+=MOD;
while(x>MOD) x-=MOD;
return x;
}
ll mult(ll a,ll b){
return (a*b)%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1) res=(res*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return res;
}
const int maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
struct segt{
int val[4*maxn],md[4*maxn];
int n;
inline void resize(int _n){
n=_n;
}
inline void mdf(int idx,int l,int r,int p,int x){
if(l>p or r<p) return;
if(l==r){
val[idx]=md[idx]=x;
return;
}
int mid=(l+r)/2;
mdf(idx*2,l,mid,p,x);
mdf(idx*2+1,mid+1,r,p,x);
val[idx]=min((ll)INF,(ll)val[idx*2]*val[idx*2+1]);
md[idx]=mult(md[idx*2],md[idx*2+1]);
}
inline int qv(int idx,int l,int r,int ql,int qr){
if(l>qr or r<ql) return 1;
if(ql<=l and r<=qr) return val[idx];
int mid=(l+r)/2;
return min((ll)INF,(ll)qv(idx*2,l,mid,ql,qr)*qv(idx*2+1,mid+1,r,ql,qr));
}
inline int qm(int idx,int l,int r,int ql,int qr){
if(l>qr or r<ql) return 1;
if(ql<=l and r<=qr) return md[idx];
int mid=(l+r)/2;
return mult(qm(idx*2,l,mid,ql,qr),qm(idx*2+1,mid+1,r,ql,qr));
}
inline ll qv(int l,int r){
return qv(1,0,n-1,l,r);
}
inline ll qm(int l,int r){
return qm(1,0,n-1,l,r);
}
inline void mdf(int p,int x){
mdf(1,0,n-1,p,x);
}
}seg;
int opt[4*maxn];
int y[maxn],n;
void build(int idx,int l,int r){
if(l==r){
opt[idx]=l;
return;
}
int mid=(l+r)/2;
build(idx*2,l,mid);
build(idx*2+1,mid+1,r);
}
void upd(int idx,int l,int r,int p){
if(r<p or l>p) return;
if(l==r){
return;
}
int mid=(l+r)/2;
upd(idx*2,l,mid,p);
upd(idx*2+1,mid+1,r,p);
if(y[opt[idx*2+1]]*seg.qv(opt[idx*2]+1,opt[idx*2+1])>=y[opt[idx*2]]) opt[idx]=opt[idx*2+1];
else opt[idx]=opt[idx*2];
}
int init(int N, int X[], int Y[]) {
n=N;
seg.resize(n);
REP(i,n) seg.mdf(i,X[i]);
build(1,0,n-1);
REP(i,n) y[i]=Y[i],upd(1,0,n-1,i);
return mult(y[opt[1]],seg.qm(0,opt[1]));
}
int updateX(int pos, int val) {
seg.mdf(pos,val);
upd(1,0,n-1,pos);
return mult(y[opt[1]],seg.qm(0,opt[1]));
}
int updateY(int pos, int val) {
y[pos]=val;
upd(1,0,n-1,pos);
return mult(y[opt[1]],seg.qm(0,opt[1]));
}
int main() {
int N;
cin>>N;
int *X = (int*)malloc(sizeof(int)*(unsigned int)N);
int *Y = (int*)malloc(sizeof(int)*(unsigned int)N);
for (int i = 0; i < N; i++) {
cin>>X[i];
}
for (int i = 0; i < N; i++) {
cin>>Y[i];
}
printf("%d\n",init(N,X,Y));
int M;
cin>>M;
for (int i = 0; i < M; i++) {
int type;
int pos;
int val;
cin>>type>>pos>>val;
if (type == 1) {
printf("%d\n",updateX(pos,val));
} else if (type == 2) {
printf("%d\n",updateY(pos,val));
}
}
return 0;
}