题目链接

洛谷
bzoj

题解

整体二分

Code

#include<bits/stdc++.h>#define LL long long
#define RG registerusing namespace std;inline int gi() {RG int x = 0; RG char c = getchar(); bool f = 0;while (c != '-' && (c < '0' || c > '9')) c = getchar();if (c == '-') c = getchar(), f = 1;while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();return f ? -x : x;
}
const int N = 50010;
int n, m;#define ls (rt<<1)
#define rs (rt<<1|1)LL t[N<<2], z[N<<2];
inline void pushdown(int rt, int l, int r) {if (z[rt]) {int mid = (l + r) >> 1;t[ls] += (mid-l+1)*z[rt]; t[rs] += (r-mid)*z[rt];z[ls] += z[rt]; z[rs] += z[rt];z[rt] = 0;}return;
}
inline void update(int rt, int l, int r, int L, int R, int k) {if (L <= l && r <= R) {z[rt] += k;t[rt] += k*(r-l+1);return ;}int mid = (l + r) >> 1;pushdown(rt, l, r);if (L <= mid)update(ls, l, mid, L, R, k);if (R > mid)update(rs, mid+1, r, L, R, k);t[rt] = t[ls]+t[rs];return ;
}inline LL query(int rt, int l, int r, int L, int R) {if (L <= l && r <= R) return t[rt];int mid = (l + r) >> 1;LL s = 0;pushdown(rt, l, r);if (L <= mid)s = query(ls, l, mid, L, R);if (R > mid)s += query(rs, mid+1, r, L, R);t[rt] = t[ls]+t[rs];return s;
}
struct Question {int op, a, b, id;LL c;
}q[N], lq[N], rq[N];
LL ans[N];void div(int l, int r, int st, int ed) {if (st > ed) return ;if (l == r) {for (int i = st; i <= ed; i++)if (q[i].op == 2)ans[q[i].id] = l;return ;}int mid = (l + r) >> 1, lt = 0, rt = 0;for (int i = st; i <= ed; i++) {if (q[i].op == 1) {if (q[i].c <= mid)lq[++lt] = q[i];else update(1, 1, n, q[i].a, q[i].b, 1), rq[++rt] = q[i];}else {LL s = query(1, 1, n, q[i].a, q[i].b);if (s >= q[i].c) rq[++rt] = q[i];else q[i].c -= s, lq[++lt] = q[i];}}for (int i = st; i <= ed; i++)if (q[i].op == 1 && q[i].c > mid) update(1, 1, n, q[i].a, q[i].b, -1);for (int i = 1; i <= lt; i++)q[st+i-1] = lq[i];for (int i = 1; i <= rt; i++)q[st+lt+i-1] = rq[i];div(l, mid, st, st+lt-1); div(mid+1, r, st+lt, ed);return ;
}int main() {n = gi(); m = gi();int k = 0;for (int i = 1; i <= m; i++) {q[i].op = gi(), q[i].a = gi(), q[i].b = gi();scanf("%lld", &q[i].c);q[i].id = (q[i].op == 2) ? ++k : 0;}div(-n, n, 1, m);for (int i = 1; i <= k; i++)printf("%lld\n", ans[i]);return 0;
}