Philosopher(set 线段树合并)
作者:互联网
- 直接维护乘积是肯定不可行的, 精度会爆炸, 于是我们来维护对数的和, 最后来计算最高位即可
- 那么转换成区间求和, 区间排序
- 区间排序的方式可以采用线段树维护最大递增块来解决,外层用set来维护线段树的区间, 然后利用线段树的合并分裂性质来操作即可
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<set>
#include<cmath>
#define ll long long
#define M (1 << 18)
#define N 20000010
#define double long double
const double eps = 1e-8;
using namespace std;
int read() {
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
double c[M], ver[M];
int num[M], n, m;
int lowbit(int x) {
return x & -x;
}
void add(int x, double v) {
for(int i = x; i <= n; i += lowbit(i)) c[i] += v;
}
double query(int x) {
double ans = 0;
for(int i = x; i; i -= lowbit(i)) ans += c[i];
return ans;
}
struct Note {
int l, r, rt, op;
Note(int ln = 0, int rn = 0, int rtn = 0, int opn = 0) {
l = ln, r = rn, rt = rtn, op = opn;
}
bool operator < (const Note &b) const {
return this->l < b.l;
}
};
#define S set<Note>::iterator
set<Note> st;
int ls[N], rs[N], cnt[N], f;
double sum[N];
void pushup(int now) {
cnt[now] = cnt[ls[now]] + cnt[rs[now]];
sum[now] = sum[ls[now]] + sum[rs[now]];
}
int merge(int x, int y) {
if(!x || !y) return x + y;
ls[x] = merge(ls[x], ls[y]);
rs[x] = merge(rs[x], rs[y]);
cnt[x] = cnt[x] + cnt[y];
sum[x] = sum[x] + sum[y];
return x;
}
S insert(Note x) {
add(x.l, sum[x.rt]);
return st.insert(x).first;
}
void Del(S it) {
add(it->l, -sum[it->rt]);
st.erase(it);
}
void split(int x, int &rt1, int &rt2, int l, int r, int k) {
rt1 = ++f;
rt2 = ++f;
if(l == r) {
cnt[rt1] = k;
sum[rt1] = ver[l] * k;
cnt[rt2] = cnt[x] - cnt[rt1];
sum[rt2] = sum[x] - sum[rt1];
return;
}
int mid = (l + r) >> 1;
if(cnt[ls[x]] >= k) {
rs[rt2] = rs[x];
split(ls[x], ls[rt1], ls[rt2], l, mid, k);
} else {
ls[rt1] = ls[x];
split(rs[x], rs[rt1], rs[rt2], mid + 1, r, k - cnt[ls[x]]);
}
pushup(rt1);
pushup(rt2);
}
S split(int x) {
if(x > n) return st.end();
S it = st.upper_bound(Note(x, 0, 0, 0));
it--;
Note hh = *it;
if(hh.l == x) return it;
int rt1, rt2;
if(!hh.op) split(hh.rt, rt1, rt2, 1, n, x - hh.l);
else split(hh.rt, rt2, rt1, 1, n, hh.r - x + 1);
Del(it);
insert(Note(hh.l, x - 1, rt1, hh.op));
return insert(Note(x, hh.r, rt2, hh.op));
}
void build(int &x, int l, int r, int k) {
x = ++f;
cnt[x]++;
sum[x] += ver[k];
if(l == r) return;
int mid = (l + r) >> 1;
if(k <= mid) build(ls[x], l, mid, k);
else build(rs[x], mid + 1, r, k);
}
int calc(int l, int r) {
S L = split(l), R = split(r + 1);
R--;
double ans = query(R->r) - query(L->l - 1);
double out = pow(10, ans - floorl(ans) + eps);
return floorl(out);
}
void updata(int l, int r, int op) {
S L = split(l);
split(r + 1);
int rt = 0;
for(S it = L; it != st.end() && (it->l) <= r; Del(it++)) {
rt = merge(rt, it->rt);
}
insert(Note(l, r, rt, op));
}
int main() {
n = read(), m = read();
for(int i = 1; i <= n; i++) num[i] = read(), ver[i] = log10(i);
for(int i = 1; i <= n; i++) {
int now;
build(now, 1, n, num[i]);
insert(Note(i, i, now, 0));
}
while(m--) {
int op = read(), l = read(), r = read();
if(op == 2) cout << calc(l, r) << "\n";
else {
op = read() ^ 1;
updata(l, r, op);
}
}
return 0;
}
标签:cnt,set,int,线段,Philosopher,rt1,ls,rt2,sum 来源: https://www.cnblogs.com/luoyibujue/p/11003496.html