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

151 2 4 6 741 52 43 44 4

 

Sample Output

Case #1:1 82 42 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logn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|1typedef 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;}