No Pain No GameTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1769Accepted 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.
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.
Output
For each test cases,for each query print the answer in one line.
Sample Input
1108 2 4 9 5 7 10 6 1 352 102 46 91 47 10
Sample Output
52243
Author
WJMZBMR
Source
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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，如果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即可。
#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|1using 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;}///求所有因子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vif(l==r) {if(tree[idx]<v)tree[idx]=v;return;}int mid=(l+r)>>1;if(x<=mid)update(lson,x,v);elseupdate(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);///预处理相同有区间的左右区间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;}
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