HDU 5726 GCD (線段樹維護區間gcd)
阿新 • • 發佈:2018-12-24
GCD
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2445 Accepted Submission(s): 855
Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,
The first line of each case contains a number N, denoting the number of integers.
The second line contains N
The third line contains a number Q, denoting the number of queries.
For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries. Output For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1).
For each query, you need to output the two numbers in a line. The first number stands for g
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
map<int,ll>mp;
const int N=1e5+7;
int n,i,j,a[N],l[N],v[N],tr[N<<2];
using namespace std;
void init(){
mp.clear();
scanf("%d",&n);
for(int i=1 ; i<=n ; i++)
scanf("%d",a+i);
for(int i=1; i <= n; i++)
for(v[i]=a[i],j=l[i]=i; j ; j=l[j]-1)
{
v[j]=__gcd(v[j],a[i]);
while(l[j] > 1 && __gcd(a[i],v[l[j]-1])==__gcd(a[i],v[j]))
l[j] = l[l[j] - 1];
mp[v[j]] += j - l[j] + 1;
}
}
void build(int l=1,int r=n,int rt=1)
{
if(l==r){
tr[rt]=a[l];
return;
}
int m=(l+r)>>1;
build(lson);
build(rson);
tr[rt]=__gcd( tr[rt<<1] , tr[rt<<1|1]);
}
int query(int L,int R,int l=1,int r=n,int rt=1)
{
if(L<=l&&r<=R) return tr[rt];
int m=(l+r)>>1;
if(R<=m) return query(L,R,lson);
if(L>m) return query(L,R,rson);
return __gcd(query(L,R,lson) , query(L,R,rson));
}
int main()
{
int t;
int x,y,k;
int cas= 1;
scanf("%d",&t);
while(t--)
{
init();
build();
scanf("%d",&k);
printf("Case #%d:\n",cas++);
while(k--)
{
scanf("%d%d",&x,&y);
int ans = query(x,y);
printf("%d %lld\n",ans,mp[ans]);
}
}
return 0;
}