HDU 5726 GCD(RMQ + 二分)GCD

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概述

GCD

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

Sample Output

  
  
   
   Case #1:
1 8
2 4
2 4
6 1



题意：给出一个数列 ，m次询问，每次询问 l，r区间内的gcd值 和 与该区间gcd值相同的区间有多少个

分析：
枚举每一个左端点，找每个左端点对应的所有gcd值区间，预处理出来，由于gcd值呈阶梯下降，所以完全可以处理，此时顺便用map统计区间个数
一开始考虑的是用线段树取gcd值，在加上二分处理，但是发现这样做其实复杂度达到了 n*logn*logn*logn ，直接T掉，然后想到用RMQ O（1）去处理gcd值，降下一个logn成功AC


   
   #include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>
//#pragma comment(linker, "/STACK:1024000000,1024000000");

using namespace std;

#define INF 0x3f3f3f3f

int n;
int a[100005];

int dp[100006][30];
int mm[100005];
void initRMQ(int n,int b[])
{
    mm[0]=-1;
    for(int i=1;i<=n;i++)
    {
        mm[i]=(i&(i-1))==0?mm[i-1]+1:mm[i-1];
        dp[i][0]=b[i];
    }
    for(int j=1;j<=mm[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 x,int y)
{
    int k=mm[y-x+1];
    return __gcd(dp[x][k],dp[y-(1<<k)+1][k]);
}

int Fr(int gcd,int i)
{
    int l=i,r=n;
    while(l+1<=r)
    {
        int mid=l+r>>1;
        int g=rmq(i,mid);
        if(__gcd(g,gcd)>=gcd) l=mid+1;
        else r=mid-1;
    }
    if(__gcd(gcd,rmq(i,l))!=gcd) l--;
    return l;
}

int main()
{
    memset(a,1,sizeof a);
    int T;
    int ca=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        initRMQ(n,a);
        map<int ,long long >mp;
        for(int i=1; i<=n; i++)
        {
            int gcd=a[i];
            int pos=i;
            while(pos<=n)
            {
                int pos1=Fr(gcd,pos);
                mp[gcd]+=pos1-pos+1;
                pos=pos1+1;
                gcd=__gcd(gcd,a[pos]);
            }
        }
        int m;
        scanf("%d",&m);
        printf("Case #%d:n",ca++);
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            int gcd=rmq(a,b);
            printf("%d %lldn",gcd,mp[gcd]);
        }
    }
    return 0;
}