HDU 5726 GCD(dp、倍增)

题意:

$给定一个N\le 10^5个数,|A_i| \le 10^9,Q\le 10^5次询问$
$定义gcd(l, r)=gcd(a_l, a_{l+1}, \cdots, a_r)$
$每次询问给定一个[l, r],查询\forall_{1\le l’\le r’\le N},gcd(l’, r’)=gcd(l, r)的(l’, r’)个数$

分析:

vector<pair<int, int> >$ f[i]:=以i结尾的区间,且gcd为first,个数为second$
$由于1个数有log级别的质因子个数,所以不同的gcd的个数也是log级别的$
$这样就可以通过dp预处理出全局的mp[g]:=gcd为g的区间个数$
$之后再通过倍增预处理出g[i][j]:=以i开始向右长度为2^j的gcd$
$对于每次询问就可以类似RMQ的做法O(1)求出gcd(l, r)了$
$总时间复杂度为O(nlog^2n)$

代码:

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//
// Created by TaoSama on 2016-07-20
// 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 <ctime>
#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 = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
typedef long long LL;
int n, a[N];
vector<pair<int, int> > f[2];
map<int, LL> mp;
int g[17][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);
clock_t _ = clock();
int t; scanf("%d", &t);
while(t--) {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
mp.clear();
int p = 0; f[p].clear();
f[p].push_back({0, 1});
for(int i = 1; i <= n; ++i) {
f[!p].clear(); f[!p].push_back({0, 1});
for(auto& u : f[p]) {
int x, cnt; tie(x, cnt) = u;
x = __gcd(x, a[i]);
bool ok = true;
for(auto& v : f[!p]) {
if(x == v.first) {
ok = false;
v.second += cnt;
break;
}
}
if(ok) f[!p].push_back({x, cnt});
// printf("%d: %d %d\n", i, x, cnt);
mp[x] += cnt; //global
}
p = !p;
}
for(int i = 1; i <= n; ++i) g[0][i] = a[i];
for(int i = 1; 1 << i <= n; ++i)
for(int j = 1; j <= n; ++j)
g[i][j] = __gcd(g[i - 1][j], g[i - 1][j + (1 << i - 1)]);
static int kase = 0;
printf("Case #%d:\n", ++kase);
int q; scanf("%d", &q);
while(q--) {
int l, r; scanf("%d%d", &l, &r);
int k = 31 - __builtin_clz(r - l + 1);
int gcd = __gcd(g[k][l], g[k][r - (1 << k) + 1]);
printf("%d %I64d\n", gcd, mp[gcd]);
}
}
#ifdef LOCAL
printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC);
#endif
return 0;
}


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