题解 CF438D 【The Child and Sequence】

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; } ```