多校 GCD

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

我们注意观察

gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)，当l固定不动的时候，r=l...nr=l...n时，我们可以容易的发现,随着rr的増大，gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)是递减的，同时gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)最多
有log\
1000,000,000log 1000,000,000个不同的值，为什么呢？因为a_{l}a​l​​最多也就有log\
1000,000,000log 1000,000,000个质因数

所以我们可以在log级别的时间处理出所有的以L开头的左区间的gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​) 那么我们就可以在n\
log\ 1000,000,000n log 1000,000,000的时间内预处理出所有的gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)然后我们可以用一个map来记录，gcd值为key的有多少个
然后我们就可以对于每个询问只需要查询对应gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)为多少，然后再在map
里面查找对应答案即可.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <algorithm>
#include <cmath>
#include <ctime>
#define LL long long

using namespace std;
int a[110000];
//pair.first为求的GCD
//pair.second为区间的R
//vec对应的下标为区间的L
vector<pair<int,int> >vec[110000];
map<int,LL>M;
int main()
{
int T,t=0;
int l,r;
int n,m;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i = 1; i <= n; ++i){
scanf("%d",&a[i]);
vec[i].clear();
}

M.clear();
for(int i = n; i >= 1; i--){
int g = 0;
/*
L+1....R3...R2....R1....R0
L+1....R3区间的GCD相同
L+1....R3+1和L...R3+2和...和L+1....R2的GCD相同
L+1....R2+1和L...R2+2和...和L+1....R1的GCD相同
L+1....R1+1和L...R1+2和...和L+1....R0的GCD相同
假设L+k<R3
__gcd(a[l],[L+1,L+k])和__gcd(a[l],[L+1,R3])相同
所以只需计算__gcd(a[l],[L+1,R3])的gcd就求出了[L,R3]的GCD了,他们相同。
以此类推...
*/
for(int j = 0; j < vec[i+1].size(); ++j){
int x = vec[i+1][j].first;
int y = vec[i+1][j].second;
int d = __gcd(x,a[i]);
if(g==d)continue;
g = d;
vec[i].push_back(make_pair(g,y));
}
//[i,i]区间也是一个GCD
if(g!=a[i])vec[i].push_back(make_pair(a[i],i));
//用map记录gcd值出现了几次
for(int j = 0; j < vec[i].size(); ++j){
//区间[i,vec[i][j].second] 的 gcd 为 vec[i][j].first
if(j+1 != vec[i].size())
M[vec[i][j].first] += (LL)(vec[i][j].second-vec[i][j+1].second);
else{
M[vec[i][j].first] += (LL)(vec[i][j].second-i+1);
}
}
}
scanf("%d",&m);
printf("Case #%d:\n",++t);
while(m--){
scanf("%d%d",&l,&r);
int x = vec[l].size()-1;
for(int i = 0; i < vec[l].size(); ++i){
if(vec[l][i].second < r){
x = i-1;
break;
}
}
printf("%d %lld\n",vec[l][x].first,M[vec[l][x].first]);
}
}
return 0;
}