分治法：整体二分

整体二分主要是把所有询问放在一起二分答案，然后把操作也一起分治

当你发现多组询问可以离线的时候

当你发现询问可以二分答案而且check复杂度对于单组询问可以接受的时候

当你发现询问的操作都是一样的的时候

BZOJ3110

修改区间，查询区间第K大

#include "cstdio"
#define lowbit(x) (x & (-x)) 
 
using namespace std;
 
const int Nmax = 50005;
 
int N, M;
struct Option{
    int sign, x, y, c;
}op[Nmax];
 
int tot = -1, ans[Nmax];
int q[Nmax], tmp[2][Nmax];
 
namespace BIT{
    int t[Nmax][2], d[Nmax][2];
 
    void update(bool s, int pos, int c)
    {
        for (int i = pos; i <= N; i += lowbit(i)) {
            if (t[i][s] != tot) { t[i][s] = tot; d[i][s] = 0; }
            d[i][s] += c;
        }
    }
     
    int get_sum(bool s, int pos)
    {
        int res = 0;
        for (int i = pos; i; i -= lowbit(i)) {
            if (t[i][s] != tot) { t[i][s] = tot; d[i][s] = 0; }
            res += d[i][s];
        }
        return res;
    }
     
    void Add(int x, int y)
    {
        update(0, x, 1); update(0, y + 1, -1);
        update(1, x, x); update(1, y + 1, -(y + 1));
    }
     
    int Query(int x, int y)
    {
        int temp = get_sum(0, y) * (y + 1) - get_sum(1, y);
        temp -= get_sum(0, x - 1) * x - get_sum(1, x - 1);
        return temp;
    }
}
 
void solve(int L, int R, int l, int r)
{
    if (L > R) return;
     
    ++tot; int mid = (l + r) >> 1;
    if (l == r) {
        for (int i = L; i <= R; ++i) if (op[q[i]].sign == 2) ans[q[i]] = mid;
        return;
    }
     
    tmp[0][0] = tmp[1][0] = 0; using namespace BIT;
    for (int i = L; i <= R; ++i) {
        int temp = q[i];
        if (op[temp].sign == 1) {
            if (op[temp].c <= mid) tmp[0][++tmp[0][0]] = temp;
            else {
                tmp[1][++tmp[1][0]] = temp;
                Add(op[temp].x, op[temp].y);
            }
        } else {
            int cnt = Query(op[temp].x, op[temp].y);
            if (cnt < op[temp].c) {
                op[temp].c -= cnt;
                tmp[0][++tmp[0][0]] = temp;
            } else tmp[1][++tmp[1][0]] = temp;
        }
    }
     
    int tl = L, t2 = L + tmp[0][0] - 1;
    for (int i = 1; i <= tmp[0][0]; ++i) q[tl++] = tmp[0][i];
    for (int i = 1; i <= tmp[1][0]; ++i) q[tl++] = tmp[1][i];
    solve(L, t2, l, mid); solve(t2 + 1, R, mid + 1, r);
}
 
int main()
{
    scanf("%d%d", &N, &M);
    for (int i = 1; i <= M; ++i) {
        scanf("%d%d%d%d", &op[i].sign, &op[i].x, &op[i].y, &op[i].c);
        q[i] = i;
    }
    solve(1, M, 1, N);
    for (int i = 1; i <= M; ++i) {
        if (op[i].sign == 2) printf("%d
", ans[i]);
    }
     
    return 0;
}