No Pain No Game
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1769 Accepted Submission(s): 748
Problem Description
Life is a game,and you lose it,so you suicide.
But you can not kill yourself before you solve this problem:
Given you a sequence of number a1, a2, ..., an.They are also a permutation of 1...n.
You need to answer some queries,each with the following format:
If we chose two number a,b (shouldn't be the same) from interval [l, r],what is the maximum gcd(a, b)? If there's no way to choose two distinct number(l=r) then the answer is zero.
But you can not kill yourself before you solve this problem:
Given you a sequence of number a1, a2, ..., an.They are also a permutation of 1...n.
You need to answer some queries,each with the following format:
If we chose two number a,b (shouldn't be the same) from interval [l, r],what is the maximum gcd(a, b)? If there's no way to choose two distinct number(l=r) then the answer is zero.
Input
First line contains a number T(T <= 5),denote the number of test cases.
Then follow T test cases.
For each test cases,the first line contains a number n(1 <= n <= 50000).
The second line contains n number a1, a2, ..., an.
The third line contains a number Q(1 <= Q <= 50000) denoting the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query.
Then follow T test cases.
For each test cases,the first line contains a number n(1 <= n <= 50000).
The second line contains n number a1, a2, ..., an.
The third line contains a number Q(1 <= Q <= 50000) denoting the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query.
Output
For each test cases,for each query print the answer in one line.
Sample Input
1
10
8 2 4 9 5 7 10 6 1 3
5
2 10
2 4
6 9
1 4
7 10
Sample Output
5
2
2
4
3
Author
WJMZBMR
Source
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5278题意： 有N个数。 是 1~N的一个排列。有M个询问， 每次询问一个区间， 问从这个区间中，取两个数的最大的最大公约数。
题解：先把查询按右区间升序排序。在将数组按顺序插入，记录当前这个数的因子出现的位置，假设之前有出现则代表这两个因子出现的
位置之间有两个数的公共约数是它，用线段树维护区间约数最大值就可以。
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define N 50050
#define lc idx<<1
#define rc idx<<1|1 using namespace std; int n,q,flag;
int a[N],tree[N*4];
int L[N],R[N];
int first[N],ans[N];
vector<int>vec; struct node {
int id;
int l,r;
} Q[N]; bool cmp(node a,node b) {
if(a.r==b.r)
return a.l<b.l;
return a.r<b.r;
} ///求全部因子
void FJ(int x) {
vec.clear();
for(int i=1; i*i<=x; i++) {
if(x%i==0) {
vec.push_back(i);
if(x/i!=i)
vec.push_back(x/i);
}
}
} void push_up(int idx) {
tree[idx]=max(tree[lc],tree[rc]);
} void build(int idx,int l,int r) {
tree[idx]=0;
if(l==r) {
return;
}
int mid=(l+r)>>1;
build(lson);
build(rson);
} void update(int idx,int l,int r,int x,int v) { //x处的值改为v
if(l==r) {
if(tree[idx]<v)
tree[idx]=v;
return;
}
int mid=(l+r)>>1;
if(x<=mid)update(lson,x,v);
else update(rson,x,v);
push_up(idx);
} int query(int idx,int l,int r,int x,int y) {
if(l>=x&&y>=r) {
return tree[idx];
}
int ans=0;
int mid=(l+r)>>1;
if(x<=mid) {
ans=max(ans,query(lson,x,y));
}
if(y>mid) {
ans=max(ans,query(rson,x,y));
}
return ans;
} void debug() {
for(int i=0; i<vec.size(); i++) {
printf("%d ",vec[i]);
}
cout<<endl;
} int main() {
//freopen("test.in","r",stdin);
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<=q; i++) {
scanf("%d%d",&Q[i].l,&Q[i].r);
Q[i].id=i;
}
sort(Q+1,Q+1+q,cmp);
memset(first,0,sizeof first);
memset(L,0,sizeof L);
memset(R,0,sizeof R);
build(1,1,n);
///预处理同样有区间的左右区间
int f=1;
L[Q[f].r]=f;
R[Q[f].r]=f;
for(int i=1; i<=q;) {
while(Q[i].r==Q[f].r&&i<=q) {
i++;
}
L[Q[f].r]=f;
R[Q[f].r]=i-1;
f=i;
}
for(int i=1; i<=n; i++) {
//FJ(a[i]);
//debug();
int xx=a[i];
for(int k=1; k*k<=xx; k++) {
if(xx%k==0) {
if(!first[k]) {
first[k]=i;
} else {
update(1,1,n,first[k],k);
first[k]=i;
}
int kk=xx/k;
if(k!=kk) {
if(!first[kk]) {
first[kk]=i;
} else {
update(1,1,n,first[kk],kk);
first[kk]=i;
}
}
}
}
int x=L[i],y=R[i];
if(x==0||y==0)continue;
for(int j=x; j<=y; j++) {
int k=Q[j].l;
if(k==i) {
ans[Q[j].id]=0;
} else {
ans[Q[j].id]=query(1,1,n,k,i);
}
}
if(y==q)break;
}
for(int i=1; i<=q; i++) {
printf("%d\n",ans[i]);
}
}
return 0;
}