HDU 5869 Different GCD Subarray Query (离线处理 树状数组)
2017-09-04 00:36
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题意:给你一个长度为n的数组,q次询问区间[L, R], 求区间内的所有子段的不同gcd值有多少种.
思路:有个知识:固定一个数为右端点,区间所能形成的gcd最多有logn个。所以可以预处理出每个数为右端点所能形
成的gcd,相同gcd取左端点靠右的。这样对询问的R排序就能离线处理,用树状数组维护即可。
代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+15;
int n, q, tree[maxn];
int a[maxn], ans[maxn], vis[maxn];
typedef pair<int, int> pp;
vector<pp> g[maxn];
struct node
{
int l, r, id;
bool operator < (const node &a) const
{
if(r == a.r) return l < a.l;
return r < a.r;
}
}qq[maxn];
int lowbit(int x)
{
return x&(-x);
}
void update(int pos, int val)
{
while(pos < maxn)
{
tree[pos] += val;
pos += lowbit(pos);
}
}
int query(int pos)
{
int res = 0;
while(pos)
{
res += tree[pos];
pos -= lowbit(pos);
}
return res;
}
void solve()
{
for(int i = 1; i <= n; i++)
{
g[i].push_back(pp(a[i], i));
int pre = a[i];
for(int j = 0; j < g[i-1].size(); j++)
{
int tmp = __gcd(a[i], g[i-1][j].first);
if(tmp != pre)
g[i].push_back(pp(tmp, g[i-1][j].second)), pre = tmp;
}
}
int cur = 0;
for(int i = 1; i <= q; i++)
{
while(cur < qq[i].r)
{
cur++;
for(int j = 0; j < g[cur].size(); j++)
{
int tmp = g[cur][j].first;
int pos = g[cur][j].second;
if(vis[tmp]) update(vis[tmp], -1);
vis[tmp] = pos;
update(vis[tmp], 1);
}
}
ans[qq[i].id] = query(qq[i].r)-query(qq[i].l-1);
}
for(int i = 1; i <= q; i++)
printf("%d\n", ans[i]);
}
int main(void)
{
while(cin >> n >> q)
{
memset(vis, 0, sizeof(vis));
memset(tree, 0, sizeof(tree));
for(int i = 0; i <= n; i++)
g[i].clear();
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= q; i++)
scanf("%d%d", &qq[i].l, &qq[i].r), qq[i].id = i;
sort(qq+1, qq+1+q);
solve();
}
return 0;
}
思路:有个知识:固定一个数为右端点,区间所能形成的gcd最多有logn个。所以可以预处理出每个数为右端点所能形
成的gcd,相同gcd取左端点靠右的。这样对询问的R排序就能离线处理,用树状数组维护即可。
代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+15;
int n, q, tree[maxn];
int a[maxn], ans[maxn], vis[maxn];
typedef pair<int, int> pp;
vector<pp> g[maxn];
struct node
{
int l, r, id;
bool operator < (const node &a) const
{
if(r == a.r) return l < a.l;
return r < a.r;
}
}qq[maxn];
int lowbit(int x)
{
return x&(-x);
}
void update(int pos, int val)
{
while(pos < maxn)
{
tree[pos] += val;
pos += lowbit(pos);
}
}
int query(int pos)
{
int res = 0;
while(pos)
{
res += tree[pos];
pos -= lowbit(pos);
}
return res;
}
void solve()
{
for(int i = 1; i <= n; i++)
{
g[i].push_back(pp(a[i], i));
int pre = a[i];
for(int j = 0; j < g[i-1].size(); j++)
{
int tmp = __gcd(a[i], g[i-1][j].first);
if(tmp != pre)
g[i].push_back(pp(tmp, g[i-1][j].second)), pre = tmp;
}
}
int cur = 0;
for(int i = 1; i <= q; i++)
{
while(cur < qq[i].r)
{
cur++;
for(int j = 0; j < g[cur].size(); j++)
{
int tmp = g[cur][j].first;
int pos = g[cur][j].second;
if(vis[tmp]) update(vis[tmp], -1);
vis[tmp] = pos;
update(vis[tmp], 1);
}
}
ans[qq[i].id] = query(qq[i].r)-query(qq[i].l-1);
}
for(int i = 1; i <= q; i++)
printf("%d\n", ans[i]);
}
int main(void)
{
while(cin >> n >> q)
{
memset(vis, 0, sizeof(vis));
memset(tree, 0, sizeof(tree));
for(int i = 0; i <= n; i++)
g[i].clear();
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= q; i++)
scanf("%d%d", &qq[i].l, &qq[i].r), qq[i].id = i;
sort(qq+1, qq+1+q);
solve();
}
return 0;
}
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