hdu 5726 GCD RMQ+二分枚举

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3641    Accepted Submission(s): 1313

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
  
  

第一个问题求区间的gcd就是一个简单的rmq，关于rmq的介绍推荐一篇博客：http://blog.youkuaiyun.com/zhangjun03402/article/details/50440136

第二问离线枚举每个点开始的每种gcd值，由于一个序列的gcd一定是递减的，而每次递减都至少减少一半，所有gcd的个数只有log（1e9），用二分来查找gcd。这样总的复杂度只有n*logn*logn。由于gcd为离散值，所以我们用map来存储。

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=110000;
int n,a[N],f[N][22];

int gcd(int x,int y)
{
    return x%y==0?y:gcd(y,x%y);
}

void rmq()
{
    for(int i=1;i<=n;i++)   f[i][0]=a[i];
    for(int j=1;j<18;j++)
    {
        for(int i=1;i<=n;i++)
        {
            if((1<<j)+i-1<=n)
            {
                f[i][j]=gcd(f[i][j-1],f[i+(1<<(j-1))][j-1]);
            }
        }
    }
}
map<int,ll>mp;

int look(int l,int r)
{
    int k=(int)log2((double)(r-l+1));
    return gcd(f[l][k],f[r-(1<<k)+1][k]);
}

void init()
{
    mp.clear();
    for(int i=1;i<=n;i++)
    {
        int g=a[i],j=i;
        while(j<=n)
        {
            g=gcd(g,a[j]);
            int l=j,r=n;
            while(l<r)
            {
                int mid=(l+r+1)>>1;
                if(look(l,mid)==g)   l=mid;
                else    r=mid-1;
            }
            mp[g]+=(l-j+1);
            j=l+1;
        }
    }
}

int main()
{
    int T,kase=0;
    cin>>T;
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)   scanf("%d",&a[i]);
        rmq();
        init();
        int q;
        scanf("%d",&q);
        printf("Case #%d:\n",++kase);
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int ans1=look(l,r);
            printf("%d %lld\n",ans1,mp[ans1]);
        }
    }
}