HDU 5726 GCD（Sparse-Table+二分）

原题链接

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

题目大意

给出一个数组a，每次询问(l,r)，问gcd(al,al+1,···ar)的值，以及在所有可能的l,r值中这个值的数目。

解题思路

参考之前求解RMQ问题的思路，dp[i][j]表示从ai开始，长度为2j的序列的gcd，这样就有递推方程

dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);

这样就可以O(nlogn)预处理，O(1)查询，接下来的问题就是如何统计出各个gcd值的数目。

首先可以枚举数列的起点，因为起点确定后，gcd的值是单调不升的，这样就可以通过二分查找查询出每个gcd有多少个值，通过map < int, long long>维护结果。

AC代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<string>
#include<queue>
#include<list>
#include<stack>
#include<set>
#include<map>
#define ll long long
#define ull unsigned long long
#define rep(i,a,b) for (int i=(a),_ed=(b);i<_ed;i++)
#define fil(a,b) memset((a),(b),sizeof(a))
#define cl(a) fil(a,0)
#define pb push_back
#define mp make_pair
#define PI 3.1415927
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define VII vector<int,int>
using namespace std;
int dp[100005][20];
int a[100005];
int gcd(int a,int b)
{
if(!b) return a;
return gcd(b,a%b);
}
int q(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1) k++;
int ans1=dp[l][k];
int ans2=dp[r-(1<<k)+1][k];
return gcd(ans1,ans2);
}
int main(void)
{
int t,n,m;
cin>>t;
for(int z=1;z<=t;++z)
{
map<int,ll> sum;
printf("Case #%d:\n",z);
scanf("%d",&n);
for(int i=1;i<=n;++i)
{
scanf("%d",&a[i]);
dp[i][0]=a[i];
}
for(int j=1;j<=17;++j)
{
for(int i=1;i+(1<<j)-1<=n;++i)
{
dp[i][j]=gcd(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
}
int ed=n+1;
for(int u=1;u<=n;++u)
{
int i=u;
while(i<=n)
{
ed=n+1;
int st=i;
while(ed>st)
{
int  mid=(st+ed)>>1;
if(q(u,i)==q(u,mid))
{
st=mid+1;
}
else
{
ed=mid;
}
}
sum[q(u,i)]+=(st-i);
i=st;

}
}
int l,r;
scanf("%d",&m);
for(int i=1;i<=m;++i)
{
scanf("%d%d",&l,&r);
int res=q(l,r);
printf("%d %I64d\n",res,sum[res]);
}
}
return 0;
}