题意
给n<1e5个娃娃,每个娃娃有属性(p),(c),(l),(r)(均在ll范围内),问你对每个娃娃(i),满足所有(l_jleq p_ileq r_j)的娃娃(j)中第(i)大的(c_i)是多少
思路
离散化后
扫描线段上的所有点,对当前点覆盖的所有线段所在的娃娃的(c_i)建权值线段树,(log)查询即可
代码
int n;
int a[maxn];
ll C[maxn];
vector<ll>v;
int find(ll x){
return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
ll P[maxn],L[maxn],R[maxn];
vector<ll>in[maxn],out[maxn],hv[maxn];
ll Padd, Pfirst, Pmod, Pprod, Cadd, Cfirst, Cmod, Cprod, Ladd, Lfirst, Lmod, Lprod, Radd, Rfirst, Rmod, Rprod;
void add(int p, int x, int l, int r, int root){
int mid = l+r>>1;
if(l==r){
a[root]+=x;return;
}
if(p<=mid)add(p,x,lson);
else add(p,x,rson);
a[root]=a[lc]+a[rc];
return;
}
int ask(int k, int l, int r, int root){
int mid = l+r>>1;
if(k>a[root])return 0;
if(l==r)return l;
if(a[rc]>=k)return ask(k,rson);
else return ask(k-a[rc],lson);
}
int main() {
scanf("%d", &n);
scanf("%lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld %lld",&Padd,&Pfirst,&Pmod,&Pprod,&Cadd,&Cfirst,&Cmod,&Cprod,&Ladd,&Lfirst,&Lmod,&Lprod,&Radd,&Rfirst,&Rmod,&Rprod);
P[1]=Pfirst%Pmod;C[1]=Cfirst%Cmod;L[1]=Lfirst%Lmod;R[1]=Rfirst%Rmod;
for(int i = 1; i <= n; i++){
if(i>1){
P[i] = (P[i-1] * Pprod + Padd + i) % Pmod;
C[i] = (C[i-1] * Cprod + Cadd + i) % Cmod;
L[i] = (L[i-1] * Lprod + Ladd + i) % Lmod;
R[i] = (R[i-1] * Rprod + Radd + i) % Rmod;
}
}
for(int i = 1; i <= n; i++){
if(L[i]>R[i])swap(L[i],R[i]);
v.pb(P[i]);v.pb(C[i]);v.pb(L[i]);v.pb(R[i]+1);
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
for(int i = 1; i <= n; i++){
in[find(L[i])].pb(find(C[i]));
out[find(R[i]+1)].pb(find(C[i]));
hv[find(P[i])].pb(i);
}
int tot = v.size();
ll sum = 0;
for(int i = 1; i <= tot; i++){
for(int j = 0; j < (int)in[i].size(); j++){
add(in[i][j],1,1,tot,1);
}
for(int j = 0; j < (int)out[i].size(); j++){
add(out[i][j],-1,1,tot,1);
}
for(int j = 0; j < (int)hv[i].size(); j++){
int now = ask(hv[i][j],1,tot,1);
if(now)sum=(sum+v[now-1])%19921228;
}
}
printf("%lld",sum%19921228);
return 0;
}