[线段树][简单复杂度分析]LOJ#6029. 「雅礼集训 2017 Day1」市场

Description

Solution

#include <bits/stdc++.h>
using namespace std;

const int N = 101010;
const long long INF = 1ll << 60;
typedef long long ll;

inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
template<typename T>
inline void read(T &x) {
static char c; x = 0; int sgn = 0;
for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
if (sgn) x = -x;
}

int a
;
ll sum[N << 2], mx[N << 2], mn[N << 2], add[N << 2];
int n, q, opt, l, r, x;

inline void PushDown(int o, int l, int r) {
if (add[o]) {
int mid = (l + r) >> 1;
add[o << 1] += add[o]; add[o << 1 | 1] += add[o];
mx[o << 1] += add[o]; mx[o << 1 | 1] += add[o];
mn[o << 1] += add[o]; mn[o << 1 | 1] += add[o];
sum[o << 1] += add[o] * (mid + 1 - l);
sum[o << 1 | 1] += add[o] * (r - mid);
add[o] = 0;
}
}
inline void PushUp(int o) {
mx[o] = max(mx[o << 1], mx[o << 1 | 1]);
mn[o] = min(mn[o << 1], mn[o << 1 | 1]);
sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
inline ll Div(ll x, ll y) {
return floor((double)x / y);
}
inline void Build(int o, int l, int r) {
add[o] = 0;
if (l == r) {
sum[o] = mx[o] = mn[o] = a[l];
return;
}
int mid = (l + r) >> 1;
Build(o << 1, l, mid);
Build(o << 1 | 1, mid + 1, r);
PushUp(o);
}
inline void Add(int o, int l, int r, int L, int R, int x) {
if (l >= L && r <= R) {
add[o] += x; mx[o] += x; mn[o] += x;
sum[o] += x * (r - l + 1);
return;
}
PushDown(o, l, r);
int mid = (l + r) >> 1;
if (L <= mid) Add(o << 1, l, mid, L, R, x);
if (R > mid) Add(o << 1 | 1, mid + 1, r, L, R, x);
PushUp(o);
}
inline void Div(int o, int l, int r, int L, int R, int d) {
if (l >= L && r <= R && mx[o] - Div(mx[o], d) == mn[o] - Div(mn[o], d)) {
ll D = Div(mx[o], d) - mx[o];
add[o] += D; mx[o] += D; mn[o] += D;
sum[o] += D * (r - l + 1);
return;
}
PushDown(o, l, r);
int mid = (l + r) >> 1;
if (L <= mid) Div(o << 1, l, mid, L, R, d);
if (R > mid) Div(o << 1 | 1, mid + 1, r, L, R, d);
PushUp(o);
}
inline ll Sum(int o, int l, int r, int L, int R) {
if (l >= L && r <= R) return sum[o];
PushDown(o, l, r);
int mid = (l + r) >> 1; ll res = 0;
if (L <= mid) res += Sum(o << 1, l, mid, L, R);
if (R > mid) res += Sum(o << 1 | 1, mid + 1, r, L, R);
return res;
}
inline ll Min(int o, int l, int r, int L, int R) {
if (l >= L && r <= R) return mn[o];
PushDown(o, l, r);
int mid = (l + r) >> 1; ll res = INF;
if (L <= mid) res = min(res, Min(o << 1, l, mid, L, R));
if (R > mid) res = min(res, Min(o << 1 | 1, mid + 1, r, L, R));
return res;
}

int main(void) {
freopen("1.in", "r", stdin);
read(n); read(q);
for (int i = 1; i <= n; i++) read(a[i]);
Build(1, 1, n);
for (int i = 1; i <= q; i++) {
read(opt); read(l); read(r); ++l; ++r;
if (opt == 1) {
read(x); Add(1, 1, n, l, r, x);
} else if (opt == 2) {
read(x); Div(1, 1, n, l, r, x);
} else if (opt == 3) {
printf("%lld\n", Min(1, 1, n, l, r));
} else {
printf("%lld\n", Sum(1, 1, n, l, r));
}
}
return 0;
}