hdu 5726 GCD(线段树+预处理)

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,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

Input

The first line of input contains a number T, which stands for the number of test cases you need to solve.

The first line of each case contains a number N, denoting the number of integers.

The second line contains N integers, a1,...,an(0<ai≤1000,000,000).

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 gcd(al,al+1,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

Sample Input

1

5

1 2 4 6 7

4

1 5

2 4

3 4

4 4

Sample Output

Case #1:

1 8

2 4

2 4

6 1

题意：给你n个数字，然后问你有多少个区间[l,r]的gcd等于询问的区间的gcd。

思路：对于查询区间的gcd，直接用线段树做就可以了。gcd有个性质，就是一个数字ai最多是log2(ai)个质因子，由于区间的gcd，随着区间的增大，gcd是单调不增的，所以对于固定的右端点ai，往左扩张，gcd的个数，不会超过log2(ai)个，所以可以直接用map存，枚举右端点，map[i]=gcd(map[i−1],ai)，因为map里最多只会有log2(ai)个值。所以可以直接暴力。

#include <iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 ll;
ll gcd(ll a,ll b)
{
if(b!=0) return gcd(b,a%b);
return a;
}
struct node
{
ll l,r,gcd;
} tree[1000005];
ll a[100005],ans,cnt;
map<ll,ll>temp1,temp2,ans1;
void build(ll id,ll l,ll r)
{
tree[id].l=l,tree[id].r=r;
if(l==r)
{
tree[id].gcd=a[l];
return;
}
ll mid=(l+r)/2;
build(2*id,l,mid);
build(2*id+1,mid+1,r);
tree[id].gcd=gcd(tree[2*id].gcd,tree[2*id+1].gcd);
}
ll Query(ll id,ll l,ll r)
{
if(tree[id].l==l&&tree[id].r==r)
return tree[id].gcd;
ll mid=(tree[id].l+tree[id].r)/2;
if(r<=mid) ans=Query(2*id,l,r);
else if(l>=mid+1) ans=Query(2*id+1,l,r);
else
{
ll t1=Query(2*id,l,mid);
ll t2=Query(2*id+1,mid+1,r);
ans=gcd(t1,t2);
}
return ans;
}
int main()
{
ll t,n,q;cnt=0;
scanf("%I64d",&t);
while(t--)
{
temp1.clear();
temp2.clear();
ans1.clear();
scanf("%I64d",&n);
for(ll i=1; i<=n; i++)
scanf("%I64d",&a[i]);
build(1,1,n);
temp1[a[1]]++,ans1[a[1]]++;
for(int i=2; i<=n; i++)
{
ll now=a[i];
temp2[now]++;
ans1[now]++;
for(auto it=temp1.begin(); it!=temp1.end(); it++)
{
int nex=gcd(now,it->first);
ans1[nex]+=it->second;
temp2[nex]+=it->second;
}
temp1.clear();
for(auto it=temp2.begin(); it!=temp2.end(); it++)
{
temp1[it->first]=it->second;
}
temp2.clear();
}
printf("Case #%I64d:\n",++cnt);
scanf("%I64d",&q);
for(ll i=1; i<=q; i++)
{
ll l,r;
ans=0;
scanf("%I64d%I64d",&l,&r);
ll ans=Query(1,l,r);
cout<<ans<<" "<<ans1[ans]<<endl;
}
}
return 0;
}