题目

Sol

二分+线段树

巧妙啊我怎么就没想到

二分答案，把数分类，大于等于\(mid\)的为\(1\)，小于的为\(0\)

相当于给\(01\)序列排序，最后判断询问位置上是不是\(1\)

线段树+lazy覆盖

# include 
    
     # 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 int Input(){    RG int 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, q, a[_], ql[_], qr[_], qo[_];int sum[_ << 2], cov[_ << 2];IL void Build(RG int x, RG int l, RG int r, RG int v){    sum[x] = 0, cov[x] = -1;    if(l == r){        sum[x] = a[l] >= v;        return;    }    RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;    Build(ls, l, mid, v), Build(rs, mid + 1, r, v);    sum[x] = sum[ls] + sum[rs];}IL void Pushdown(RG int x, RG int l, RG int mid, RG int r){    RG int ls = x << 1, rs = x << 1 | 1;    cov[ls] = cov[rs] = cov[x];    sum[ls] = (mid - l + 1) * cov[x];    sum[rs] = (r - mid) * cov[x];    cov[x] = -1;}IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){    if(L <= l && R >= r){        sum[x] = (r - l + 1) * v;        cov[x] = v;        return;    }    RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;    if(cov[x] != -1) Pushdown(x, l, mid, r);    if(L <= mid) Modify(ls, l, mid, L, R, v);    if(R > mid) Modify(rs, mid + 1, r, L, R, v);    sum[x] = sum[ls] + sum[rs];}IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){    if(cov[x] != -1) return cov[x] * (R - L + 1);    if(L <= l && R >= r) return sum[x];    RG int mid = (l + r) >> 1;    if(R <= mid) return Query(x << 1, l, mid, L, R);    if(L > mid) return Query(x << 1 | 1, mid + 1, r, L, R);    return Query(x << 1, l, mid, L, mid) + Query(x << 1 | 1, mid + 1, r, mid + 1, R);}IL bool Check(RG int mid){    Build(1, 1, n, mid);    for(RG int i = 1; i <= m; ++i){        RG int cnt = Query(1, 1, n, ql[i], qr[i]);        if(!cnt || cnt == qr[i] - ql[i] + 1) continue;        if(qo[i]){            Modify(1, 1, n, ql[i], ql[i] + cnt - 1, 1);            Modify(1, 1, n, ql[i] + cnt, qr[i], 0);        }        else{            Modify(1, 1, n, ql[i], qr[i] - cnt, 0);            Modify(1, 1, n, qr[i] - cnt + 1, qr[i], 1);        }    }    return Query(1, 1, n, q, q);}int main(RG int argc, RG char* argv[]){    n = Input(), m = Input();    for(RG int i = 1; i <= n; ++i) a[i] = Input();    for(RG int i = 1; i <= m; ++i) qo[i] = Input(), ql[i] = Input(), qr[i] = Input();    q = Input();    RG int l = 1, r = n, ans = 0;    while(l <= r){        RG int mid = (l + r) >> 1;        if(Check(mid)) ans = mid, l = mid + 1;        else r = mid - 1;    }    printf("%d\n", ans);    return 0;}