HDU 5726 GCD RMQ

GCD

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

Total Submission(s): 681    Accepted Submission(s): 207


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

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

 

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题意：给1e5个数，1e5个查询，查询 [L,R]区间所有数的gcd，以及这个gcd在整个a数组中有多少个（区间的gcd） == 这个gcd。

区间gcd，这个直接把以前线段树模版求和改成两个数的gcd就行。但是我tle了，改成RMQ，改进查询的复杂度。RMQ也是把求和改成求两个数gcd就行。

重点是求所有区间gcd个数。暴力枚举n方复杂度不行，可以发现一个区间内，越往后gcd是呈阶梯式减小的。最多减小log1000,000,000次。可以枚举每个数，二分查询递减的时候每个阶梯的位置，然后用一个map记录下+=这个阶梯gcd的长度。

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e5+10;

int n;
LL dp
[25];   ///以i开头的后面长度为2的j次方的gcd
LL a
;        ///用于输入的数组
map<int,LL> mp; ///存gcd个数

LL gcd(LL a,LL b){
return b == 0?a:gcd(b,a%b);
}

int ask(int l,int r)   ///RMQ查询
{
int x=(int)(log(r-l+1)/log(2));
return gcd(dp[l][x],dp[r-(1<<x)+1][x]);
}

void ST()               ///RMQ
{
for(int i = 1;i <= n;i++)
{
dp[i][0] = a[i];
}
for(int j = 1;j <= (int)(log(n)/log(2));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 main(void)
{
int T;
scanf("%d",&T);
int times = 1;
while(T--){
mp.clear();
scanf("%d",&n);
for(int i = 1;i <= n;i++) scanf("%I64d",&a[i]);
ST();
for(int i = 1;i <= n;i++){   ///预处理区间gcd的个数
int val = a[i];
int pos = i;
while(pos <= n){
val = ask(i,pos);
int l = pos,r = n;
while(l <= r){
int mid = (l+r)/2;
if(ask(i,mid) == val) l = mid+1;
else                  r = mid-1;
}
mp[val] += r-pos+1;
pos = l;
}
}
printf("Case #%d:\n",times++);
int qq;
scanf("%d",&qq);
while(qq--){
int a,b;
scanf("%d%d",&a,&b);
LL tmp = ask(a,b);
printf("%I64d %I64d\n",tmp,mp[tmp]);
}
}
return 0;
}