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hdu 5381 The sum of gcd(线段树+gcd)

时间:2018-03-27 21:07:25      阅读:217      评论:0      收藏:0      [点我收藏+]

标签:shu   algo   track   cto   std   oid   vector   int   cstring   

题目链接:hdu 5381 The sum of gcd


将查询离线处理,依照r排序,然后从左向右处理每一个A[i],碰到查询时处理。用线段树维护。每一个节点表示从[l,i]中以l为起始的区间gcd总和。所以每次改动时须要处理[1,i-1]与i的gcd值。可是由于gcd值是递减的,成log级,对于每一个gcd值记录其区间就可以。然后用线段树段改动,可是是改动一个等差数列。


#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e4 + 5;
#define lson(x) (x<<1)
#define rson(x) ((x<<1)|1)

namespace SegTree {
	int lc[maxn << 2], rc[maxn << 2];
	ll A[maxn << 2], D[maxn << 2], S[maxn << 2];

	void maintain (int u, ll a, ll d) {
		A[u] += a;
		D[u] += d;
		int n = rc[u] - lc[u] + 1;
		S[u] += a * n + d * (n-1) * n / 2;
	}

	void pushup(int u) {
		S[u] = S[lson(u)] + S[rson(u)];
	}

	void pushdown (int u) {
		if (A[u] || D[u]) {
			int mid = ((lc[u] + rc[u]) >> 1) + 1;
			maintain(lson(u), A[u], D[u]);
			maintain(rson(u), A[u] + (mid - lc[u]) * D[u], D[u]);
			A[u] = D[u] = 0;
		}
	}

	void build (int u, int l, int r) {
		lc[u] = l;
		rc[u] = r;
		A[u] = D[u] = S[u] = 0;

		if (l == r) {
			return;
		}

		int mid = (l + r) >> 1;
		build (lson(u), l, mid);
		build (rson(u), mid + 1, r);
		pushup(u);
	}

	void modify (int u, int l, int r, ll a, ll d) {
		if (l <= lc[u] && rc[u] <= r) {
			maintain(u, a + (lc[u] - l) * d, d);
			return;
		}

		pushdown(u);
		int mid = (lc[u] + rc[u]) >> 1;
		if (l <= mid)
			modify(lson(u), l, r, a, d);
		if (r > mid)
			modify(rson(u), l, r, a, d);
		pushup(u);
	}

	ll query (int u, int l, int r) {
		if (l <= lc[u] && rc[u] <= r)
			return S[u];

		pushdown(u);
		ll ret = 0;
		int mid = (lc[u] + rc[u]) >> 1;
		if (l <= mid)
			ret += query(lson(u), l, r);
		if (r > mid)
			ret += query(rson(u), l, r);
		return ret;
	}
};

int gcd (int a, int b) {
	return b == 0 ?

a : gcd(b, a%b); } struct State { int l, r, idx; State (int l = 0, int r = 0, int idx = 0): l(l), r(r), idx(idx) {} bool operator < (const State& u) const { return r < u.r; } }S[maxn], G[maxn]; int N, Q, A[maxn]; ll R[maxn]; void init () { scanf("%d", &N); for (int i = 1; i <= N; i++) scanf("%d", &A[i]); scanf("%d", &Q); for (int i = 1; i <= Q; i++) { scanf("%d%d", &S[i].l, &S[i].r); S[i].idx = i; } sort(S + 1, S + 1 + Q); SegTree::build(1, 1, N); } void solve () { int n = 0, p = 1; for (int i = 1; i <= N; i++) { for (int j = 0; j < n; j++) G[j].idx = gcd(G[j].idx, A[i]); G[n++] = State(i, i, A[i]); int mv = 0; for (int j = 1; j < n; j++) { if (G[mv].idx == G[j].idx) G[mv] = State(G[mv].l, G[j].r, G[j].idx); else G[++mv] = G[j]; } n = mv + 1; //printf("%d:%d\n", i, n); for (int i = 0; i < n; i++) { // printf("%d %d %d\n", G[i].l, G[i].r, G[i].idx); int k = G[i].r - G[i].l + 1; SegTree::modify(1, G[i].l, G[i].r, 1LL * G[i].idx * k, -G[i].idx); if (G[i].l > 1) SegTree::modify(1, 1, G[i].l - 1, 1LL * G[i].idx * k, 0); } while (p <= N && S[p].r == i) { R[S[p].idx] = SegTree::query(1, S[p].l, S[p].l); p++; } } } int main () { int cas; scanf("%d", &cas); while (cas--) { init (); solve (); for (int i = 1; i <= Q; i++) printf("%lld\n", R[i]); } return 0; }



hdu 5381 The sum of gcd(线段树+gcd)

标签:shu   algo   track   cto   std   oid   vector   int   cstring   

原文地址:https://www.cnblogs.com/zhchoutai/p/8659410.html

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