GCD (hdu 5726)

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1497    Accepted Submission(s): 483

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

 

 

Author

HIT

 

 

Source

2016 Multi-University Training Contest 1

思路：RMQ+二分

一开始我用的线段树去维护各个区间的gcd但是超时，并且我没有优化。

暴力统计的各个区间的gcd；

后来发现gcd的性质，也就是求数的gcd，这些数的个数越多那么gcd越小，所以从左到右，以某个端点开始的gcd的大小随着区间增大而减小，那么二分统计以某个点为端点的gcd

最多每个端点二分30次。

然后还是超时。

然后改用RMQ基于稀疏表的，查询O（n）；

复杂度（n*log(n)）;

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<map>
#include<math.h>
using namespace std;
typedef long long LL;
int RMQ[20][100005];
int ans[100005];
map<int,LL>my;
int gcd(int n,int m)
{
        if(m==0)
                return n;
        else  if(n%m==0)
        {
                return m;
        }
        else return gcd(m,n%m);
}
int main(void)
{
        int i,j,k;
        int n,m;
        scanf("%d",&k);
        int ca=0;
        while(k--)
        {
                my.clear();
                scanf("%d",&n);
                for(i=1; i<=n; i++)
                {
                        scanf("%d",&ans[i]);
                }
                for(i=1; i<=n; i++)
                {
                        RMQ[0][i]=ans[i];
                }
                scanf("%d",&m);
                for(i=1; i<20; i++)
                {
                        for(j=1; j<=n; j++)
                        {
                                if(j+(1<<i)-1<=n)
                                {
                                        RMQ[i][j]=gcd(RMQ[i-1][j],RMQ[i-1][j+(1<<(i-1))]);
                                }
                        }
                }
                for(i=1; i<=n; i++)
                {
                        for(j=i; j<=n;)
                        {
                                int t=log2(j-i+1);
                                int ac=gcd(RMQ[t][i],RMQ[t][j-(1<<t)+1]);
                                int l=j;
                                int r=n;
                                int id=0;
                                while(l<=r)
                                {
                                        int mid=(l+r)/2;
                                        int c=log2(mid-i+1);
                                        int ak=gcd(RMQ[c][i],RMQ[c][mid-(1<<c)+1]);
                                        if(ak>=ac)
                                        {
                                                id=mid;
                                                l=mid+1;
                                        }
                                        else r=mid-1;
                                }
                                my[ac]+=id-j+1;
                                j=id+1;
                        }
                }
                printf("Case #%d:\n",++ca);
                while(m--)
                {
                        int x,y;
                        scanf("%d %d",&x,&y);
                        if(x>y)
                        {
                            swap(x,y);
                        }
                        int ct=log2(y-x+1);
                        int acc=gcd(RMQ[ct][x],RMQ[ct][y-(1<<ct)+1]);
                        printf("%d %lld\n",acc,my[acc]);
                }
        }
        return 0;
}