L Euler Function
题解
线段树维护区间欧拉函数和。
代码
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#include <bits/stdc++.h> #define lc rt << 1 #define rc rt << 1 | 1 using namespace std; typedef long long LL; const int N = 100009; const int M = 109; const LL Mod = 998244353; LL p[M]; class SegmentTree { public: void build(int l, int r, int rt) { t[rt].val = t[rt].lzy = 1LL; t[rt].tag.reset(); if(l>=r) return; int mid = (l + r) >> 1; build(l, mid, lc); build(mid + 1, r, rc); pushup(rt); } void update(int l, int r, int rt, int L, int R, int ip, int k) { if(l > r || r < L || l > R || L > R) return; if(L <= l && r <= R && t[rt].tag[ip]) { for (int i = 1; i <= k; i++) { (t[rt].val *= p[ip]) %= Mod; (t[rt].lzy *= p[ip]) %= Mod; } return; } if(l==r) { (t[rt].val *= p[ip] - 1LL) %= Mod; (t[rt].lzy *= p[ip] - 1LL) %= Mod; t[rt].tag[ip] = 1; for (int i = 1; i < k; i++) { (t[rt].val *= p[ip]) %= Mod; (t[rt].lzy *= p[ip]) %= Mod; } return; } int mid = (l + r) >> 1; pushdown(rt); update(l, mid, lc, L, R, ip, k); update(mid + 1, r, rc, L, R, ip, k); pushup(rt); return; } LL query(int l, int r, int rt, int L, int R) { if(r < L || l > R || l > r || L > R) return 0; if(L <= l && r <= R) return t[rt].val % Mod; int mid = (l + r) >> 1; pushdown(rt); LL lval = query(l, mid, lc, L, R); LL rval = query(mid + 1, r, rc, L, R); pushup(rt); return (lval + rval) % Mod; } private: struct Node { LL val, lzy; bitset<30> tag; }; Node t[4 * N]; void pushdown(int rt) { (t[lc].val *= t[rt].lzy) %= Mod; (t[lc].lzy *= t[rt].lzy) %= Mod; (t[rc].val *= t[rt].lzy) %= Mod; (t[rc].lzy *= t[rt].lzy) %= Mod; t[rt].lzy = 1LL; } void pushup(int rt) { t[rt].val = (t[lc].val + t[rc].val) % Mod; t[rt].tag = t[lc].tag & t[rc].tag; } }; SegmentTree t; bitset<30> g[M]; int cnt, n, m; void init() { for (int i = 2; i <= 100; i++) { bool flag = 1; for (int j = 2; j * j <= i; j++) { if(i % j) continue; flag = 0; break; } if(!flag) continue; p[++cnt] = i; } } int main() { #ifndef ONLINE_JUDGE freopen("input.in", "r", stdin); freopen("output.out", "w", stdout); #endif init(); scanf("%d%d", &n, &m); t.build(1, n, 1); for (int i = 1; i <= n; i++) { int x; scanf("%d", &x); for (int j = 1; j <= cnt; j++) { int tmp = 0; while(x % p[j] == 0) tmp++, x /= p[j]; if(tmp) t.update(1, n, 1, i, i, j, tmp); } } for (int i = 1; i <= m; i++) { int op; scanf("%d", &op); if(op == 0) { int l, r, x; scanf("%d%d%d", &l, &r, &x); for (int j = 1; j <= cnt; j++) { int tmp = 0; while(x % p[j] == 0) tmp++, x /= p[j]; if(tmp) t.update(1, n, 1, l, r, j, tmp); } } else { int l, r; scanf("%d%d", &l, &r); LL ans = t.query(1, n, 1, l, r); printf("%lld\n", ans); } } return 0; } |
叨叨几句... NOTHING