HDU 5726 GCD(RMQ + 二分)

GCD

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).

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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).

 

[align=left]Sample Input[/align]

1
5
1 2 4 6 7
4
1 5
2 4
3 4
4 4

 

[align=left]Sample Output[/align]

Case #1:
1 8
2 4
2 4
6 1

题意：给出一个数列 ，m次询问，每次询问 l，r区间内的gcd值 和 与该区间gcd值相同的区间有多少个

分析：
枚举每一个左端点，找每个左端点对应的所有gcd值区间，预处理出来，由于gcd值呈阶梯下降，所以完全可以处理，此时顺便用map统计区间个数
一开始考虑的是用线段树取gcd值，在加上二分处理，但是发现这样做其实复杂度达到了 n*logn*logn*logn ，直接T掉，然后想到用RMQ O（1）去处理gcd值，降下一个logn成功AC

#include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>
//#pragma comment(linker, "/STACK:1024000000,1024000000");

using namespace std;

#define INF 0x3f3f3f3f

int n;
int a[100005];

int dp[100006][30];
int mm[100005];
void initRMQ(int n,int b[])
{
mm[0]=-1;
for(int i=1;i<=n;i++)
{
mm[i]=(i&(i-1))==0?mm[i-1]+1:mm[i-1];
dp[i][0]=b[i];
}
for(int j=1;j<=mm
;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 rmq(int x,int y)
{
int k=mm[y-x+1];
return __gcd(dp[x][k],dp[y-(1<<k)+1][k]);
}

int Fr(int gcd,int i)
{
int l=i,r=n;
while(l+1<=r)
{
int mid=l+r>>1;
int g=rmq(i,mid);
if(__gcd(g,gcd)>=gcd) l=mid+1;
else r=mid-1;
}
if(__gcd(gcd,rmq(i,l))!=gcd) l--;
return l;
}

int main()
{
memset(a,1,sizeof a);
int T;
int ca=1;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
initRMQ(n,a);
map<int ,long long >mp;
for(int i=1; i<=n; i++)
{
int gcd=a[i];
int pos=i;
while(pos<=n)
{
int pos1=Fr(gcd,pos);
mp[gcd]+=pos1-pos+1;
pos=pos1+1;
gcd=__gcd(gcd,a[pos]);
}
}
int m;
scanf("%d",&m);
printf("Case #%d:\n",ca++);
while(m--)
{
int a,b;
scanf("%d%d",&a,&b);
int gcd=rmq(a,b);
printf("%d %lld\n",gcd,mp[gcd]);
}
}
return 0;
}