BZOJ 2161 布娃娃(权值线段树)

#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$l_j\leq p_i\leq r_j$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;
}