hdu 5726 gcd rmq 求相同gcd的区间数量

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：给一个数组a，大小为n，接下来有m个询问，每次询问给出l、r，定义f[l,r]=gcd（al，al+1，…，ar），问f[l,r]的值 和 有多少对（l’，r’）使得f[l’,r’]=f[l,r]。n<=10万，m<=10万，1<=l<=r<=n，1<=l’<=r’。

思路：

　　第一步比较简单，预处理一下，定义f[i][j]为：ai开始，连续2^j个数的最大公约数，所以f[1][0]=a[1],f[1][1]=gcd(a1,a2),f[1][2]=gcd(a1,a2,a3,a4)。其实就是动态规划，让i从1-n，让j从0-17，递推上去即可。

　　递推公式如下：

　　1. f[i][0]=a[i];

　　2. f[i][j]=gcd(f[i][j-1],f[i+(1 << j-1)][j-1]);

　　就如同f[1][2]=gcd(f[1][1],f[3][1])=gcd(gcd(f[1][0],f[2][0]),gcd(f[3][0],f[4][0]));

　　通过上述预处理，查询时就只需O(logn)时间，如下：

　　令k=log2(r-l+1)，look(l,r)=gcd(f[l][k],f[r-(1 << k)+1][k]);

　　注：f[l][k] 和 f[r-(1 << k)+1][k]可能会有重叠，但不影响最终的gcd值。

　　比赛时第二步没想出来，太可惜了。。。

　　第二步，我们可以枚举左端点 i 从1-n，对每个i，二分右端点，计算每种gcd值的数量，因为如果左端点固定，gcd值随着右端点的往右，呈现单调不增，而且gcd值每次变化，至少除以2，所以gcd的数量为nlog2(n)种，可以开map < int,long long > 存每种gcd值的数量，注意n大小为10万，所以数量有可能爆int。

#include <bits/stdc++.h>using namespace std;typedef long long ll;int n;int gd[100100][18];map<int,long long> mp;int gcd(int a,int b){    return (b==0)?a:gcd(b,a%b);}int a[100100];void ST(){    for(int j=1;(1<<j)<=n;j++)    {        for(int i=1;i+(1<<j)-1<=n;i++)        {            gd[i][j]=gcd(gd[i][j-1],gd[i+(1<<(j-1))][j-1]);        }    }}int rmq(int i,int j){    int k=0;    while((1<<(k+1))<=j-i+1) k++;    return gcd(gd[i][k],gd[j-(1<<k)+1][k]);}int main(){    int t;    scanf("%d",&t);    for(int cc=1;cc<=t;cc++)    {        mp.clear();        printf("Case #%d:\n",cc );        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            gd[i][0]=a[i];        }        ST();        for(int i=1;i<=n;i++)        {            int l=i,r=n;            int g=a[i];            while(l<=n)            {            int ll=l;            while(l<=r){                int mid=(l+r)>>1;                if(rmq(i,mid)==g)                    l=mid+1;                else r=mid-1;            }            mp[g]+=(r-ll+1);                g=gcd(g,a[l]);              r=n;            }           }        int q;        scanf("%d",&q);        while(q--){            int c,d;            scanf("%d%d",&c,&d);            int res=rmq(c,d);            printf("%d %lld\n",res,mp[res]);        }    }}