# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
4648 | cki86201 | Evaluation (kriii1_E) | C++98 | 0 ms | 13972 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,ll> P;
#define Fi first
#define Se second
const ll M=1e9+7;
ll T[1<<19];
int C[1<<19],down[1<<19];
int in[200020][2],N;
int p[200020],ord[200020];
int X[200020],W[200020];
ll ans,now;
void Mod(ll &x)
{
if(x>0)x = x%M;
else if(x%M==0)x = 0;
else x = M-x%M;
}
void Mod(int &x)
{
if(x>0)x = x%M;
else if(x%M==0)x = 0;
else x = M-x%M;
}
bool comp0(const int &a,const int &b){return in[a][0]!=in[b][0]?in[a][0]<in[b][0]:a<b;}
bool comp1(const int &a,const int &b){return in[a][1]!=in[b][1]?in[a][1]<in[b][1]:a<b;}
void build(int s,int e,int t)
{
if(s==e){
down[t] = X[s];
return;
}
int m = (s+e)>>1;
build(s,m,t<<1);
build(m+1,e,t<<1|1);
down[t] = down[t<<1] + down[t<<1|1];
}
P update(int l,int r,int s,int e,int t,int u)
{
P d=P(0,0);
if(l<=s&&e<=r){
d = P(down[t],T[t]);
C[t] += u;
T[t] += (ll)u*down[t];
Mod(T[t]);
return d;
}
int m = (s+e)>>1;
if(l<=m){
P tmp = update(l,r,s,m,t<<1,u);
d.Fi += tmp.Fi, d.Se += tmp.Se;
}
if(r>m){
P tmp = update(l,r,m+1,e,t<<1|1,u);
d.Fi += tmp.Fi, d.Se += tmp.Se;
}
d.Se += (ll)d.Fi * C[t];
T[t] += (ll)d.Fi * u;
Mod(d.Fi), Mod(d.Se), Mod(T[t]);
return d;
}
int main()
{
scanf("%d",&N);
int i;
for(i=0;i<N;i++){
scanf("%d%d%d%d%d",in[2*i],in[2*i]+1,in[2*i+1],in[2*i+1]+1,p+2*i);
p[2*i+1]=p[2*i];
in[2*i+1][0]++,in[2*i+1][1]++;
ord[2*i]=2*i,ord[2*i+1]=2*i+1;
}
sort(ord,ord+2*N,comp0);
for(i=0;i<2*N;i++){
if(i)X[i]=in[ord[i]][0]-in[ord[i-1]][0];
W[ord[i]]=i;
}
sort(ord,ord+2*N,comp1);
build(1,2*N-1,1);
for(i=0;i<2*N;i++){
int x = ord[i]>>1;
int a = W[2*x]+1, b = W[2*x+1], f = p[ord[i]]*((ord[i]&1)?-1:1);
if(i)ans += now * (in[ord[i]][1] - in[ord[i-1]][1]);
now += (ll)(in[2*x+1][0]-in[2*x][0])*f*f;
int u=(f+M)%M;
now += 2LL*u*update(a,b,1,2*N-1,1,f).Se;
Mod(now);
}
printf("%d",int(ans%M));
}
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---|---|---|---|---|
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