题面
Sol
整体二分
比较经典，练手题
每次的修改会影响一个区间，我用的是线段树覆盖
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, ans[_], vis[_], tag[_ << 2];
ll sum[_ << 2], tmp[_];
struct Data{
int l, r, c, id;
} q[_], q1[_], q2[_];
IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
sum[x] += (R - L + 1) * v;
if(L == l && R == r){
tag[x] += v;
return;
}
RG int mid = (l + r) >> 1;
if(R <= mid) Modify(x << 1, l, mid, L, R, v);
else if(L > mid) Modify(x << 1 | 1, mid + 1, r, L, R, v);
else Modify(x << 1, l, mid, L, mid, v), Modify(x << 1 | 1, mid + 1, r, mid + 1, R, v);
}
IL ll Query(RG int x, RG int l, RG int r, RG int L, RG int R, RG ll ad){
if(L == l && R == r) return sum[x] + ad * (r - l + 1);
RG int mid = (l + r) >> 1; ad += tag[x];
if(R <= mid) return Query(x << 1, l, mid, L, R, ad);
if(L > mid) return Query(x << 1 | 1, mid + 1, r, L, R, ad);
return Query(x << 1, l, mid, L, mid, ad) + Query(x << 1 | 1, mid + 1, r, mid + 1, R, ad);
}
IL void Solve(RG int l, RG int r, RG int L, RG int R){
if(L > R) return;
if(l == r){
for(RG int i = L; i <= R; ++i) ans[q[i].id] = l;
return;
}
RG int mid = (l + r) >> 1, t1 = 0, t2 = 0;
for(RG int i = L; i <= R; ++i)
if(vis[q[i].id]) tmp[q[i].id] = Query(1, 1, n, q[i].l, q[i].r, 0);
else if(q[i].c > mid) Modify(1, 1, n, q[i].l, q[i].r, 1);
for(RG int i = L; i <= R; ++i)
if(vis[q[i].id]){
if(tmp[q[i].id] >= q[i].c) q2[++t2] = q[i];
else q[i].c -= tmp[q[i].id], q1[++t1] = q[i];
}
else{
if(q[i].c <= mid) q1[++t1] = q[i];
else q2[++t2] = q[i];
}
for(RG int i = L; i <= R; ++i)
if(!vis[q[i].id] && q[i].c > mid) Modify(1, 1, n, q[i].l, q[i].r, -1);
for(RG int i = L, j = 1; j <= t1; ++i, ++j) q[i] = q1[j];
for(RG int i = L + t1, j = 1; j <= t2; ++i, ++j) q[i] = q2[j];
Solve(l, mid, L, L + t1 - 1); Solve(mid + 1, r, L + t1, R);
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input();
RG int mn = 2147483647, mx = -mn;
for(RG int i = 1; i <= m; ++i){
RG int op = Input(); q[i].l = Input(), q[i].r = Input(), q[i].c = Input(), q[i].id = i;
if(op == 1) mx = max(mx, q[i].c), mn = min(mn, q[i].c);
vis[i] = op == 2;
}
Solve(mn, mx, 1, m);
for(RG int i = 1; i <= m; ++i)
if(vis[i]) printf("%d\n", ans[i]);
return 0;
}