HDU 4630 No Pain No Game(离线思想、BIT)

题意:

$N\le 5\times 10^4,Q\le 5\times 10^4,给定一个1\sim N的排列,Q次询问$
$每次询问[L,R]区间任意2个数的gcd的最大值,规定1个数答案是0$

分析:

$经典离线套路题了,询问右端点排序$
$从左往右扫描每个数A_i(相当于固定了右端点)$
$对于每个数的约数x,维护它的倍数中的前一个位置为last_x$
$用BIT维护,只要单点更新所有约数x的last_x为x$
$发现所有以i为右端点的询问[L,i]只要查询[L,i]的最大值就好了$
$这里其实把BIT倒过来就可以完成了,向前更新,向后查询$
$筛法预处理1\sim N的约数,总时间复杂度为O(nlogn)$

代码:

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//
// Created by TaoSama on 2016-03-22
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 5e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
struct BIT {
int n, b[N];
void init(int _n) {
n = _n;
memset(b, 0, sizeof b);
}
void add(int i, int v) {
for(; i; i -= i & -i) b[i] = max(b[i], v);
}
int sum(int i) {
int ret = 0;
for(; i <= n; i += i & -i) ret = max(ret, b[i]);
return ret;
}
} bit;
vector<int> divisors[N];
vector<pair<int, int> > qs[N];
int n, q, a[N], ans[N];
int last[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
for(int i = 1; i <= 5e4; ++i)
for(int j = i; j <= 5e4; j += i)
divisors[j].push_back(i);
int t; scanf("%d", &t);
while(t--) {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
scanf("%d", &q);
for(int i = 1; i <= n; ++i) qs[i].clear();
for(int i = 1; i <= q; ++i) {
int l, r; scanf("%d%d", &l, &r);
qs[r].push_back({l, i});
}
bit.init(n);
memset(last, 0, sizeof last);
for(int i = 1; i <= n; ++i) {
int v = a[i];
for(int x : divisors[v]) {
if(last[x]) bit.add(last[x], x);
last[x] = i;
}
for(auto q : qs[i]) ans[q.second] = bit.sum(q.first);
}
for(int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}
return 0;
}


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