[HDU 5726]GCD

Time Limit: 10000/5000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)

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

题意：
给一个长度为N数列，有M组询问，每组询问有两个数$l,r$返回两个答案，
（1）$gcd(a[i])$ $(i∈[l,r])$
（2）有多少对$[l',r']$ 满足区间的最大公约数等于第一问

题解
第一问可以用rmq解决
那么问题是第二问，第二问我们可以经过一个预处理，处理出区间内所有可能出现的每个最大公因数所对应的区间个数。这个可以通过枚举区间左端点来解决，枚举区间左端点，然后以此为左端点有着连续、相同最大公因数的区间的右端点肯定是连续的，可以通过二分来确定他们的位置然后用map储存对应的最大公因数的区间个数就可以了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<cstdlib>
#include<cstring>
#define LiangJiaJun main
#define ll long long
using namespace std;
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
int n,m,a[100004];
map<int , ll >mert;
int gp[100004][24],mm[100004];
int pre(){
    mm[0]=-1;
    for(int i=1;i<=n;i++){
        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
        gp[i][0]=a[i];
    }
    for(int j=1;j<=mm[n];j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            gp[i][j]=gcd(gp[i][j-1],gp[i+(1<<(j-1))][j-1]);
        }
    }
    return 0;
}
int ask(int x,int y){
    int k=mm[y-x+1];
    return gcd(gp[x][k],gp[y-(1<<k)+1][k]);
}

int w33ha(int CASE){
    mert.clear();
    memset(gp,0,sizeof(gp));
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    pre();
    for(int i=1;i<=n;i++){
        int mid,l,r,now=a[i],rb=i;
        while(rb<=n){
            l=rb;r=n;int cpr=-1;
            while(l<=r){
                mid=(l+r)>>1;
                if(ask(i,mid)==now){
                    cpr=mid;l=mid+1;
                }
                else r=mid-1;
            }
            mert[now]+=cpr-rb+1;
            rb=cpr+1;
            now=gcd(now,a[rb]);
        }
    }
    printf("Case #%d:\n",CASE);
    scanf("%d",&m);
    for(int i=1;i<=m;i++){
        int l,r;scanf("%d%d",&l,&r);
        int g=ask(l,r);
        printf("%d %lld\n",g,mert[g]);
    }
    return 0;
}
int LiangJiaJun (){
    int T;scanf("%d",&T);
    for(int i=1;i<=T;i++)w33ha(i);
    return 0;
}