2019 CCPC wannfly winter camp Day 5

C - Division

思路：我们考虑到一点，从大往小取得顺序是不会有问题的，所以可以直接主席树，但是开不下空间，我们可以log分段求。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;

int n, q, all, hs[2 * N], hcnt, a[N], bin[31];
LL ans[5 * N], sum[N], csum[N];
pair<PII, int> qus[5 * N];
vector<int> vc[30][N];

int root[2 * N], tot;

struct node {
    LL sum;
    int cnt, ls, rs;
} o[2 * N * 20];

void update(int p, int l, int r, int &x, int y) {
    x = ++tot;
    o[x] = o[y];
    o[x].sum += p;
    o[x].cnt++;
    if(l == r) return;
    int mid = l + r >> 1;
    if(p <= hs[mid]) update(p, l, mid, o[x].ls, o[y].ls);
    else update(p, mid + 1, r, o[x].rs, o[y].rs);
}

LL query(int res, int l, int r, int x, int y) {
    if(o[x].cnt - o[y].cnt <= res) return o[x].sum - o[y].sum;
    if(l == r) return 1ll * res * hs[l];
    int mid = l + r >> 1, cntr = o[o[x].rs].cnt - o[o[y].rs].cnt;
    if(cntr >= res) return query(res, mid + 1, r, o[x].rs, o[y].rs);
    else return o[o[x].rs].sum - o[o[y].rs].sum + query(res-cntr, l, mid, o[x].ls, o[y].ls);
}

int main() {
    for(int i = bin[0] = 1; i <= 30; i++) bin[i] = bin[i - 1] << 1;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), sum[i] = sum[i - 1] + a[i];
    for(int i = 1; i <= q; i++) {
        scanf("%d%d%d", &qus[i].fi.fi, &qus[i].fi.se, &qus[i].se);
        ans[i] = sum[qus[i].fi.se] - sum[qus[i].fi.fi - 1];
    }
    for(int i = 1, j = 29; i <= n; i++, j = 29) {
        while(a[i]) {
            int val = a[i] - a[i] / 2;
            while(val < bin[j]) j--;
            vc[j][i].push_back(val);
            a[i] /= 2;
        }
    }
    for(int k = 29; k >= 0; k--) {
        tot = 0; hcnt = 0; all = 0;
        for(int i = 1; i <= n; i++) csum[i] = csum[i - 1] + SZ(vc[k][i]);
        if(!csum[n]) continue;
        for(int i = 1; i <= n; i++) for(auto& t : vc[k][i]) hs[++hcnt] = t;
        sort(hs + 1, hs + 1 + hcnt); hcnt = unique(hs + 1, hs + 1 + hcnt) - hs - 1;
        for(int i = 1; i <= n; i++) for(auto& t : vc[k][i]) update(t, 1, hcnt, root[all + 1], root[all]), all++;
        for(int i = 1; i <= q; i++) {
            if(!qus[i].se) continue;
            int L = qus[i].fi.fi, R = qus[i].fi.se, has = csum[R] - csum[L - 1];
            if(has >= qus[i].se) {
                ans[i] -= query(qus[i].se, 1, hcnt, root[csum[R]], root[csum[L - 1]]);
                qus[i].se = 0;
            } else {
                ans[i] -= query(has, 1, hcnt, root[csum[R]], root[csum[L - 1]]);
                qus[i].se -= has;
            }
        }
    }
    for(int i = 1; i <= q; i++) printf("%lld
", ans[i]);
    return 0;
}

/*
*/