Philosopher(set 线段树合并)
Posted luoyibujue
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- 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;
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