hdu 5726 RMQ+二分

最新推荐文章于 2019-03-18 20:48:45 发布

原创最新推荐文章于 2019-03-18 20:48:45 发布 · 402 阅读

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本文介绍了一种高效解决区间GCD查询及计数问题的方法，通过预处理和数据结构优化，实现了快速查询指定区间内的最大公约数及其相同GCD值区间的数量。

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GCD

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 513 Accepted Submission(s): 164

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

Author

HIT

Source

2016 Multi-University Training Contest 1

题意：

这是2016多校第一场的1004题，当时没写出来，题意就是求给定区间的gcd值以及与gcd为该值的区间的个数。

思路：

随着区间的扩大，区间gcd值一定是减小的，在减小的情况下，每次至少要减小一个因数，因此最多只会有因数个区间内gcd值是不同的。然后具体处理就是了，具体看代码。

代码：

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

#define INF 0x3f3f3f3f
const int MAXN =100010;
const int mod=1000007;
int dp[MAXN][20];
int mm[MAXN];
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 n,q;
int a[MAXN];
struct node
{
    int d;
    long long cnt;
};
list<node> lis[mod];
long long & get(int d)
{
    int h=d%mod;
    for(list<node>::iterator it=lis[h].begin();it!=lis[h].end();it++)
    {
        if((*it).d==d) return (*it).cnt;
    }
    lis[h].push_front((node){d,0});
    for(list<node>::iterator it=lis[h].begin();it!=lis[h].end();it++)
    {
        if((*it).d==d) return (*it).cnt;
    }
}
int half_find(int s,int l,int r,int d)
{
    int mid;
    while(l<=r)
    {

        mid=(l+r)>>1;
        if(rmq(s,mid)>=d) l=mid+1;
        else r=mid-1;
    }
    return l;
}
int main()
{
    int t,tt=0;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<mod;i++) lis[i].clear();
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        initRMQ(n,a);
        for(int i=1;i<=n;i++)
        {
            int l,r=i,d=a[i];
            while(r<=n)
            {
                l=r;
                r=n;
                r=half_find(i,l,r,d);

                long long& cnt=get(d);
                cnt+=r-l;
                d=rmq(i,r);
            }
        }
        printf("Case #%d:\n",++tt);
        scanf("%d",&q);
        for(int i=1;i<=q;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            int d=rmq(x,y);
            long long cnt=get(d);
            printf("%d %lld\n",d,cnt);
        }
    }
    return 0;
}