BZOJ3932: [CQOI2015]任务查询系统

传送门
题目大意：在[s,e]（闭区间）上加入一个数字k（就相当于在s时刻放入，e+1时刻取出）。支持询问在t时刻时的第k大数。

想一想就可以知道主席树能水过这道题。

/**************************************************************
    Problem: 3932
    User: geng4512
    Language: C++
    Result: Accepted
    Time:3808 ms
    Memory:192900 kb
****************************************************************/

#include<cstdio>
#include<vector>
#include<algorithm>
#define MAXN 100005
#define LL long long
using namespace std;
struct node {
    int lc, rc, cnt;
    long long sum;
} t[MAXN*80];
vector<int> L[MAXN], R[MAXN];
int n, m, rt[MAXN], cnt, tag, mark[MAXN*80];
char c;
inline void GET(int &n) {
    n = 0;
    do c = getchar(); while('0' > c || c > '9');
    do n = n*10+c-'0', c = getchar(); while('0' <= c && c <= '9');
}
inline void Build(int &rt, int x) {
    t[rt = ++ cnt] = t[x];
    mark[rt] = tag;
}
void Insert(int &rt, int p, int v, int l, int r, char f) {
    if(mark[p] != tag) Build(rt, p);
    t[rt].cnt += f;
    t[rt].sum += v*f;
    if(l >= r) return;
    int mid = (l + r) >> 1;
    if(v <= mid) Insert(t[rt].lc, t[p].lc, v, l, mid, f);
    else Insert(t[rt].rc, t[p].rc, v, mid+1, r, f);
}
long long Query(int x, int l, int r, int k) {
    if(l >= r) return 1LL * k * l;
    int mid = (l + r) >> 1;
    if(t[t[x].lc].cnt >= k) return Query(t[x].lc, l, mid, k);
    return Query(t[x].rc, mid+1, r, k - t[t[x].lc].cnt) + t[t[x].lc].sum;
}
int main() {
    int s, e, v, x, mn=MAXN, mx=-1;
    long long p = 1;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++ i) {
        GET(s); GET(e); GET(v);
        L[s].push_back(v); R[e].push_back(v);
        mn = min(mn, s); mx = max(mx, e);
    }
    for(int i = mn; i <= mx; ++ i) {
        Build(rt[i], rt[i-1]); ++tag;
        for(int j = 0; j < (int)L[i].size(); ++ j)
            Insert(rt[i], rt[i], L[i][j], 1, 1e7, 1);
        for(int j = 0; j < (int) R[i-1].size(); ++ j)
            Insert(rt[i], rt[i], R[i-1][j], 1, 1e7, -1);
    }
    for(int i = 1; i <= m; ++ i) {
        GET(x); GET(s); GET(e); GET(v);
        v = 1 + (p*s+e)%v;
        if(v >= t[rt[x]].cnt) p = t[rt[x]].sum;
        else p = Query(rt[x], 1, 1e7, v);
        printf("%lld\n", p);
    }
    return 0;
}