hdu 5726(二分)

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1233    Accepted Submission(s): 382


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

/*
难点在于计数，关键要发现gcd的递减性
对于1到n的(l,r)
对于固定的l
gcd(l,r)>=gcd(l,r+1)
对于固定的r
gcd(l,r)<=gcd(l+1,r)
因此可以不用逐一地进行计数
方法为：
枚举每一个左区间,对满足gcd(l,ri)的右区间进行二分查找
跳跃着进行计数
*/
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <map>
#include <vector>
#define scan1(x)     scanf("%d",&x)
#define scan2(x,y)   scanf("%d%d",&x,&y)
#define scan3(x,y,z) scanf("%d%d%d",&x,&y,&z)
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const int Max=1e6+10;
int dp[Max][30];
map<int,LL> vis;
int A[Max];
int gcd(int x,int y)
{
    if(x<y) swap(x,y);
    return (y==0?x:gcd(y,x%y));
}
void RMQ_init(int n)
{
    for(int i=1; i<=n; i++) dp[i][0]=A[i];
    for(int j=1; (1<<j)<=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 RMQ(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}
void Init()
{
   vis.clear();
}
int L[Max],R[Max];
int main()
{
    int T,ca=1;
    for(scan1(T); T; T--)
    {
        int n,m,num;
        scan1(n);
        for(int i=1; i<=n; i++) scan1(A[i]);
        RMQ_init(n);
        Init();
        scan1(m);
        for(int i=1; i<=m; i++)
        {
            scan2(L[i],R[i]);
            num=RMQ(L[i],R[i]);
            vis.insert(make_pair(num,0));
        }
        int l,r,mid,d1,d2,ans,nex;
        for(int i=1; i<=n; i++)
        {
            nex=i;
            while(nex<=n)
            {
                d1=RMQ(i,nex);
                l=nex;r=n;
                while(l<=r)
                {
                    mid=(l+r)>>1;
                    d2=RMQ(i,mid);
                    if(d2>=d1) l=mid+1,ans=mid;
                    else r=mid-1;
                }
                if(vis.find(d1)!=vis.end())
                vis[d1]+=(ans-nex)+1;
                nex=r+1;
            }
        }
        printf("Case #%d:\n",ca++);
        for(int i=1; i<=m; i++)
        {
            ans=RMQ(L[i],R[i]);
            printf("%d %lld\n",ans,vis[ans]);
        }
    }
    return 0;
}