题解 CF438D 【The Child and Sequence】
HomuraCat
2018-10-26 17:46:12
用线段树维护
问题是区间取膜怎么搞呀,取起来超级慢的QAQ
我们可以发现,每个数取模后如果它改变了,那么它必然缩小至少二分之一。
比如说一个$a\%b=c\quad(b<a)$,我们有$a=kb+c$且$b>c,k>1$,那么就有$a>2c$
有了这点就好办了,我们维护一个区间最大值,每次取膜的时候判断一下最大值如果小于膜数的话就不处理了。
这样每个数最多被膜$loga_i$次,总复杂度$O(nlognloga_i)$
代码:
```cpp
#include<bits/stdc++.h>
#define fo(i, a, b) for (int i = (a); i <= (b); ++i)
#define fd(i, a, b) for (int i = (a); i >= (b); --i)
#define edge(i, u) for (int i = head[u], v = e[i].v; i; i = e[i].nxt, v = e[i].v)
#define N 100005
#define pb push_back
#define F first
#define S second
#define ll long long
#define inf 1000000007
#define mp std::make_pair
#define lowbit(x) (x & -x)
#define ls (k << 1)
#define rs (k << 1 | 1)
int n, m, a[N], x, y, pos, val, opt, tot, l, r;
struct node{
ll sum, max, add;
}t[N << 2];
inline void pushdown (int k)
{
if (t[k].add)
{
t[ls].add += t[k].add;
t[rs].add += t[k].add;
t[k].sum += t[k].add;
t[k].add = 0;
}
}
inline void build (int k, int l, int r)
{
if (l == r) {t[k].max = t[k].sum = a[l]; return;}
int mid = l + r >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
t[k].sum = t[ls].sum + t[rs].sum;
t[k].max = std::max(t[ls].max, t[rs].max);
}
inline void modify (int k, int l, int r)
{
if (l == r && l == pos) {t[k].max = t[k].sum = val; return;}
pushdown(k);
int mid = l + r >> 1;
if (pos <= mid) modify(ls, l, mid);
else modify(rs, mid + 1, r);
t[k].sum = t[ls].sum + t[rs].sum;
t[k].max = std::max(t[ls].max, t[rs].max);
}
inline ll query (int k, int l, int r, int x, int y)
{
if (x <= l && r <= y) {return t[k].sum + t[k].add;}
pushdown(k);
int mid = l + r >> 1;
if (y <= mid) return query(ls, l, mid, x, y);
if (x > mid) return query(rs, mid + 1, r, x, y);
return query(ls, l, mid, x, y) + query(rs, mid + 1, r, x, y);
}
inline void modulo (int k, int l, int r, int x, int y)
{
if (x <= l && r <= y && t[k].max < val) return;
if (l == r) {t[k].sum = t[k].max = t[k].sum % val; return;}
pushdown(k);
int mid = l + r >> 1;
if (x <= mid) modulo(ls, l, mid, x, y);
if (mid < y) modulo(rs, mid + 1, r, x, y);
t[k].sum = t[ls].sum + t[rs].sum;
t[k].max = std::max(t[ls].max, t[rs].max);
}
int main ()
{
scanf("%d %d", &n, &m);
fo (i, 1, n)
scanf("%d", &a[i]);
build(1, 1, n);
fo (I, 1, m)
{
scanf("%d", &opt);
if (opt == 1)
{
scanf("%d %d", &l, &r);
printf("%lld\n", query(1, 1, n, l, r));
}
else
if (opt == 2)
{
scanf("%d %d %d", &l, &r, &val);
modulo(1, 1, n, l, r);
}
else
{
scanf("%d %d", &pos, &val);
modify(1, 1, n);
}
}
return 0;
}
```