GCD HDU - 5726 (ST算法+二分）

最新推荐文章于 2022-06-24 12:37:59 发布

原创最新推荐文章于 2022-06-24 12:37:59 发布 · 766 阅读

0 ·

CC 4.0 BY-SA版权

二分同时被 2 个专栏收录

16 篇文章

订阅专栏

st算法

3 篇文章

订阅专栏

本文介绍了一种解决区间GCD查询问题的方法，并通过预处理实现快速查询，同时统计区间内具有相同GCD值的子区间的数量。采用RMQ算法进行区间最大公约数的计算，使用map来统计相同GCD值的出现频率。

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,rl,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 NN, denoting the number of integers.

The second line contains NN 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,rili,ri, stand for the i-th queries.
Output
For each case, you need to output “Case #:t” at the beginning.(with quotes, tt 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

想到lST了，但是第二步统计数量没想好。

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
#include<map>
#define N 100005
using namespace std;
int dp[N][18];
int n;
int gcd(int a,int b)
{
    return a%b==0?b:gcd(b,a%b);
}
void RMQ()
{
    int cnt=ceil(log(n*1.0)/log(2.0));

    for(int i=1;i<=cnt;i++)
    {
        for(int j=1;j<=n;j++)
        {
            dp[j][i]=dp[j][i-1];
            if(j+(1<<(i-1))<=n)
                dp[j][i]=gcd(dp[j][i],dp[j+(1<<(i-1))][i-1]);
        }
    }
}
int getGCD(int l,int r)
{
    int cnt=(int)log2((double)(r-l+1));
    return gcd(dp[l][cnt],dp[r-(1<<cnt)+1][cnt]);
}
 map<int,long long> mp;
void init()
{
    mp.clear();
    for(int i=1;i<=n;i++)
    {
        int g=dp[i][0],j=i;
        while(j<=n)
        {
            int l=j,r=n;
            while(l<r)
            {
                int mid=(l+r+1)>>1;
                if(getGCD(i,mid)==g) l=mid;
                else r=mid-1;
            }
            mp[g]+=l-j+1;
            j=l+1;
            g=getGCD(i,j);
        }
    }
}
int main()
{
    //cout<<log2(100000);
    int t;
    int l,r;
    scanf("%d",&t);
    int ca=1;
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&dp[i][0]);
        RMQ();
        init();
        //cout<<"jaja";
        int q;
        scanf("%d",&q);
        printf("Case #%d:\n",ca++);
        while(q--)
        {
            scanf("%d%d",&l,&r);
            int gcdd=getGCD(l,r);
            printf("%d %lld\n",gcdd,mp[gcdd]);
        }
    }
    return 0;
}