hdu5726 GCD(gcd +二分+rmq)

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，m个询问，每一个询问都是一个区间，让你先计算出这段区间所有数的gcd，然后问1~n所有连续区间中gcd的值等于询问区间gcd的区间个数。

思路：考虑到如果固定区间左端点L，那么右端点从L+1变化到n的过程中gcd最多变化log(区间内最大的数的大小)次（因为每次变化至少除以2）,那么我们就可以枚举左端点，然后每次二分值连续的区间，然后都存到map里就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lson th<<1
#define rson th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define Key_value ch[ch[root][1]][0]
map<int,ll>mp;
map<int,ll>::iterator it;

int q[100100][2];

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int gcd1[100100][30];

int a[100006];
void init_rmq(int n)
{
    int i,j;
    for(i=1;i<=n;i++){
        gcd1[i][0]=a[i];
    }

    for(j=1;j<=20;j++){
        for(i=1;i<=n;i++){
            if(i+(1<<j)-1<=n){
                gcd1[i][j]=gcd(gcd1[i][j-1],gcd1[i+(1<<(j-1))][j-1]);
                gcd1[i][j]=gcd(gcd1[i][j-1],gcd1[i+(1<<(j-1))][j-1]);
            }
        }
    }
}

int getgcd(int l,int r)
{
    int k,i;
    if(l>r)swap(l,r);
    k=(log((r-l+1)*1.0)/log(2.0));
    return gcd(gcd1[l][k],gcd1[r-(1<<k)+1][k]);
}

int main()
{
    int n,m,i,j,T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        mp.clear();
        init_rmq(n);
        int l,r,mid;
        for(i=1;i<=n;i++){
            //printf("----->%d\n",i);
            int val=a[i];
            int pos=i;
            while(pos<=n){
                val=getgcd(i,pos);
                l=pos,r=n;
                while(l<=r){
                    mid=(l+r)/2;
                    if(getgcd(i,mid)==val)l=mid+1;
                    else r=mid-1;
                }
                mp[val]+=(r-pos+1);
                pos=l;
            }

        }
        scanf("%d",&m);
        for(i=1;i<=m;i++){
            scanf("%d%d",&q[i][0],&q[i][1]);
        }
        printf("Case #%d:\n",++cas);
        for(i=1;i<=m;i++){
            printf("%d %lld\n",getgcd(q[i][0],q[i][1]),mp[getgcd(q[i][0] ,q[i][1] ) ] );
        }
    }
    return 0;
}