GCD(st表+二分)

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3432    Accepted Submission(s): 1227

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:第一行T代表测试组数,第二行一个整数 N 代表长为 N 的序列，再一行是一个整数 Q ，代表有 Q 次询问，

然后 Q 行,每行两个整数 l,r 输出区间[l,r]的gcd 和有多少个区间 gcd 等于 gcd[l,r]

 

题解:想了很久，没想到好办法，后来发现RMQ这种问题，线段树还是不行的，如果离线的话，st表查询时间复杂度比线段树更小，为O(1)

st算法是用来求解给定区间RMQ的最值

st表学习:http://www.cnblogs.com/yangjiyuan/p/5493274.html

学了st表后，查询有多少区间gcd==gcd[l,r]还是不知道怎么搞，题解是,用二分离线预处理...

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;

#define LL long long
#define MAXN 100005

int n;
int num[MAXN];
int mn[MAXN];
int dp[MAXN][20];
map<int,LL> res;

int gcd(int a,int b)
{   return b==0?a:gcd(b,a%b);}

void Init()
{
    mn[0]=-1;
    for (int i=1;i<=n;i++)
    {
        mn[i]=((i&(i-1))==0)?mn[i-1]+1:mn[i-1];
        dp[i][0]=num[i];
    }
    for (int j=1;j<=mn[n];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 st_calc(int l,int r)
{
    int k = mn[r-l+1];
    return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}

void erfen()
{
    res.clear();
    for (int i=1;i<=n;i++)  //以i为左起点的区间的所有公约数
    {
        int cur = i , gc = num[i];  //用二分求出
        while (cur <= n)
        {
            int l=cur,r=n;
            while(l<r)
            {
                int mid = (l+r+1)>>1;
                if (st_calc(i,mid)==gc) l=mid;
                else r = mid -1;
            }
            if (res.count(gc)) res[gc]+=l-cur+1;
            else res[gc]=l-cur+1;
            cur = l + 1;
            if (cur<=n)
                gc = gcd(gc,num[cur]);
        }
    }
}

int main()
{
    int t;
    int cas=0;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        Init();

        erfen();

        int q;
        scanf("%d",&q);
        printf("Case #%d:\n",++cas);
        while (q--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int kk = st_calc(x,y);
            printf("%d %lld\n",kk,res[kk]);
        }
    }
    return 0;
}