整体二分模板题
如果没有修改的询问区间第k小or大，一般把原始值看成赋值操作，这样可以把询问和赋值同时二分，正确性显然。
如果是单点修改，同样可以用树状数组赋值，修改一个数看-1再+1，因为每次增加和修改是成对出现的且二分不改变询问和修改的顺序，所以显然二分也是正确的。
区间插入的话，和普通的整体二分差不多，改成线段树维护区间小于mid的数的个数，把增加一个数看成是线段树上区间的加+1。
// luogu-judger-enable-o2
#include
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
using LL = long long;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)){
w |= ch == '-', ch = getchar();
}
while(isdigit(ch)){
ret = (ret << 3) + (ret << 1) + (ch ^ 48);
ch = getchar();
}
return w ? -ret : ret;
}
template
inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }
template
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
const int N = 50005;
int n, m, cur, cnt, ans[N], lazy[N<<2], cls[N<<2];
LL sum[N<<2];
struct Query{
int op, id, x, y;
LL k;
Query(){}
Query(int x, int y, LL k): x(x), y(y), k(k){
op = 1, id = 0;
}
Query(int op, int id, int x, int y, LL k): op(op), id(id), x(x), y(y), k(k){}
}v[N<<1], lq[N<<1], rq[N<<1];
void push_up(int rt){
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void push_down(int rt, int l, int r){
if(cls[rt]){
cls[rt] = 0;
lazy[rt << 1] = lazy[rt << 1 | 1] = 0;
sum[rt << 1] = sum[rt << 1 | 1] = 0;
cls[rt << 1] = cls[rt << 1 | 1] = 1;
}
if(lazy[rt]){
int mid = (l + r) >> 1;
sum[rt << 1] += lazy[rt] * (mid - l + 1);
sum[rt << 1 | 1] += lazy[rt] * (r - mid);
lazy[rt << 1] += lazy[rt], lazy[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
}
}
void insert(int rt, int l, int r, int il, int ir, int val){
if(l == il && r == ir){
sum[rt] += (r - l + 1) * val;
lazy[rt] += val;
return;
}
push_down(rt, l, r);
int mid = (l + r) >> 1;
if(ir <= mid) insert(rt << 1, l, mid, il, ir, val);
else if(il > mid) insert(rt << 1 | 1, mid + 1, r, il, ir, val);
else insert(rt << 1, l, mid, il, mid, val), insert(rt << 1 | 1, mid + 1, r, mid + 1, ir, val);
push_up(rt);
}
LL query(int rt, int l, int r, int ql, int qr){
if(l == ql && r == qr){
return sum[rt];
}
push_down(rt, l, r);
int mid = (l + r) >> 1;
if(qr <= mid) return query(rt << 1, l, mid, ql, qr);
else if(ql > mid) return query(rt << 1 | 1, mid + 1, r, ql, qr);
return query(rt << 1, l, mid, ql, mid) + query(rt << 1 | 1, mid + 1, r, mid + 1, qr);
}
void solve(int L, int R, int l, int r){
if(l > r) return;
if(L == R){
for(int i = l; i <= r; i ++){
if(v[i].op == 2) ans[v[i].id] = L;
}
return;
}
int mid = (L + R) >> 1;
int lp = 0, rp = 0;
for(int i = l; i <= r; i ++){
if(v[i].op == 1){
if(v[i].k > mid) insert(1, 1, n, v[i].x, v[i].y, 1), rq[++rp] = v[i];
else lq[++lp] = v[i];
}
else{
LL ret = query(1, 1, n, v[i].x, v[i].y);
if(ret >= v[i].k) rq[++rp] = v[i];
else v[i].k -= ret, lq[++lp] = v[i];
}
}
cls[1] = 1, sum[1] = lazy[1] = 0;
for(int i = 1; i <= lp; i ++) v[l + i - 1] = lq[i];
for(int i = 1; i <= rp; i ++) v[l + lp + i - 1] = rq[i];
solve(L, mid, l, l + lp - 1);
solve(mid + 1, R, l + lp, r);
}
int main(){
//freopen("data.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i ++){
int opt, x, y; LL k;
scanf("%d%d%d%lld", &opt, &x, &y, &k);
if(opt == 1){
v[++cur] = Query(x, y, k);
}
else{
v[++cur] = Query(2, ++ cnt, x, y, k);
}
}
solve(0, n, 1, cur);
for(int i = 1; i <= cnt; i ++){
printf("%d\n", ans[i]);
}
return 0;
}