两道离线线段树。
比赛时候没想到。。。。
扫描数组，i从1到n，线段树维护从1到i每一个约数（1~50000）的出现的最近位置，线段树存储的是约数的最大值
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int N=50050;
struct Interval{
int l,r,id;
}in[N];
struct Tree{
int l,r,Max;
}tree[N<<2];
int n,m;
int a[N];
int pre[N];
int ans[N];
vector<int> factor[N];
void init(){
for(int i=1;i<N;i++)
for(int j=i;j<N;j+=i)
factor[j].push_back(i);
}
bool cmp(struct Interval a,struct Interval b){
return a.r<b.r;
}
void build(int l,int r,int id){
tree[id].l=l;
tree[id].r=r;
tree[id].Max=0;
if(l!=r){
int mid=(l+r)>>1;
build(l,mid,id<<1);
build(mid+1,r,id<<1|1);
}
}
void update(int pos,int val,int id){
tree[id].Max=max(tree[id].Max,val);
if(tree[id].l==tree[id].r) return ;
int mid=(tree[id].l+tree[id].r)>>1;
if(mid>=pos) update(pos,val,id<<1);
else update(pos,val,id<<1|1);
}
int query(int l,int r,int id){
if(tree[id].l==l && tree[id].r==r) return tree[id].Max;
int mid=(tree[id].l+tree[id].r)>>1;
if(mid>=r) return query(l,r,id<<1);
else if(mid<l) return query(l,r,id<<1|1);
else return max(query(l,mid,id<<1),query(mid+1,r,id<<1|1));
}
int main(){
init();
int t,T;
int i,j,k;
scanf("%d",&T);
for(t=1;t<=T;t++){
scanf("%d",&n);
build(1,n,1);
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(i=0;i<m;i++){
scanf("%d %d",&in[i].l,&in[i].r);
in[i].id=i;
}
sort(in,in+m,cmp);
memset(pre,0,sizeof(pre));
for(i=1,j=0;i<=n && j<m;i++){
for(k=0;k<factor[a[i]].size();k++){
int tem=factor[a[i]][k];
if(pre[tem]!=0){
update(pre[tem],tem,1);
}
pre[tem]=i;
}
while(j<m && in[j].r==i){
ans[in[j].id]=query(in[j].l,in[j].r,1);
j++;
}
}
for(i=0;i<m;i++) printf("%d
",ans[i]);
}
return 0;
}
BOJ的那题找不到了，做完这道题之后，发现这两题很像，都是离线线段树做法。（这道题还是队友出的，膜拜）
先为数组中每个点i一个最小区间l,r满足a[i]>a[l] && a[i]<a[r],l<i<r
然后i从1开始扫描，每次把能加入的点加入，然后处理右端点为i的查询
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
const int MAXN = 100000+5;
const int INF = 0x3f3f3f3f;
int T, N, Q, a[MAXN], s[MAXN], t[MAXN];
int idq[MAXN], q[MAXN], cnt[MAXN], p[MAXN];
int x[MAXN], y[MAXN], ida[MAXN];
int Tr[MAXN<<2], mark[MAXN<<2];
void Build(int idx, int L, int R)
{
Tr[idx] = mark[idx] = 0;
if (L == R)
return;
int left = idx<<1, right = idx<<1|1, mid = (L+R)>>1;
Build(left, L, mid);
Build(right, mid+1, R);
}
void PushDown(int idx)
{
int left = idx<<1, right = idx<<1|1, &mk = mark[idx];
Tr[left] += mk;
mark[left] += mk;
Tr[right] += mk;
mark[right] += mk;
mk = 0;
}
void Update(int idx, int L, int R, int l, int r, int c)
{
if (l <= L && R <= r)
{
Tr[idx] += c;
mark[idx] += c;
return;
}
if (mark[idx])
PushDown(idx);
int left = idx<<1, right = idx<<1|1, mid = (L+R)>>1;
if (l <= mid)
Update(left, L, mid, l, r, c);
if (mid < r)
Update(right, mid+1, R, l, r, c);
}
int Query(int idx, int L, int R, int x)
{
if (x == L && R == x)
return Tr[idx];
if (mark[idx])
PushDown(idx);
int left = idx<<1, right = idx<<1|1, mid = (L+R)>>1;
if (x <= mid)
return Query(left, L, mid, x);
else
return Query(right, mid+1, R, x);
}
bool cmpq(const int &a, const int &b)
{
return t[a] < t[b];
}
bool cmpa(const int &a, const int &b)
{
return y[a] < y[b];
}
int main()
{
//freopen("data.in", "r", stdin);
//freopen("data.out", "w", stdout);
scanf("%d", &T);
for (int cas = 1; cas <= T; cas++)
{
printf("Case %d:
", cas);
scanf("%d", &N);
stack<int> stal, star;
a[0] = 0;
stal.push(0);
for (int i = 1; i <= N; i++)
{
scanf("%d", &a[i]);
while (a[i] <= a[stal.top()])
stal.pop();
x[i] = stal.top();
stal.push(i);
}
a[N+1] = INF;
star.push(N+1);
for (int i = N; i >= 1; i--)
{
while (a[i] >= a[star.top()])
star.pop();
y[i] = star.top();
star.push(i);
ida[i] = i;
}
sort(ida+1, ida+1+N, cmpa);
scanf("%d", &Q);
for (int i = 1; i <= Q; i++)
{
scanf("%d%d", &s[i], &t[i]);
if (s[i] > t[i])
swap(s[i], t[i]);
idq[i] = i;
q[i] = 0;
}
sort(idq+1, idq+1+Q, cmpq);
Build(1, 1, N);
for (int i = 1, j = 1, k = 1; i <= N; i++)
{
for (; y[ida[k]] == i && k <= N; k++) if (x[ida[k]])
Update(1, 1, N, 1, x[ida[k]], 1);
for (; t[idq[j]] == i && j <= Q; j++)
q[idq[j]] = Query(1, 1, N, s[idq[j]]);
}
for (int i = 1; i <= Q; i++)
printf("%d
", q[i]);
}
return 0;
}