HDU 5726 GCD [ST表+暴力二分]【数据结构|杂类】

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=5726

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GCD

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

Total Submission(s): 3164 Accepted Submission(s): 1146

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

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题目大意:

有一个长度为N的序列,Q次查询,每次查询需求解出gcd{A},A={ai|i∈[l,r]}及与区间gcd值相同的区间有多少个.

解题思路

首先对与区间gcd 我们可以用线段树或者ST表预处理下,因为没有元素更改,相比之下后者更好(复杂度上,代码量上)

然后就是如何就解与区间gcd值相同的区间有多少个.

这个我们只能枚举一端,然后向右找,然后用个map来存储结果,这样的话复杂度是O(n2),显然不可取.

但是对于一个区间上的gcd来说一定是不增的.这样就有了单调性,我们就可以二分做了,但这样需要三层循环解决.,而且在枚举不同端点的时候还会有重复统计的情况.

然后考虑,一个gcd的递减的过程最多也就这能递减因子数个,那么我们每一次记录的就是一种gcd值的区间,然后接下来串过来就好了,

最后的复杂度就是O(n∗因子个数∗二分)

附本题代码

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#include <bits/stdc++.h>

using namespace std;

#define INF        (~(1<<31))
#define INFLL      (~(1ll<<63))
#define pb         push_back
#define mp         make_pair
#define abs(a)     ((a)>0?(a):-(a))
#define lalal      puts("*******");
#define s1(x)      scanf("%d",&x)
#define Rep(a,b,c) for(int a=(b);a<=(c);a++)
#define Per(a,b,c) for(int a=(b);a>=(c);a--)
#define no         puts("NO")

typedef long long int LL ;
typedef unsigned long long int uLL ;

const int    N   = 100000+7;
const int    MOD = 1e9+7;
const double eps = 1e-6;
const double PI  = acos(-1.0);

inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
void fre(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
}
inline int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}
/***********************************************************************/
int n;
int a
;
int st
[20],mm
;
void ST(){
for(int j=1;(1<<j)<=n; j++)
for(int i=1; i+(1<<j)-1<=n; i++)
st[i][j]=gcd(st[i][j-1],st[i+(1<<(j-1))][j-1]);
}
void initrmp(int x){
mm[0]=-1;
for(int i=1;i<=x;i++)    mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
}

int getST(int l,int r){
int k=mm[r-l+1];
return gcd(st[l][k],st[r-(1<<k)+1][k]);
}
map<int ,LL> ans;
int main(){
initrmp(100000);
int _,kcase;
while(~scanf("%d",&_)){
kcase = 0;
while(_--){
ans.clear();
n=read();
for(int i=1;i<=n;i++)    a[i]=read(),st[i][0]=a[i];
ST();
for(int i=1;i<=n;i++){
int cur = i , gc = a[i];
while( cur <= n ){
int l = cur , r = n;
while( l < r ){
int mid = l + r + 1 >> 1;
if(getST(i,mid)==gc) l = mid ;
else r = mid - 1;
}
if(ans.count(gc)) ans[gc] +=(l-cur+1);
else ans[gc]=(l-cur+1);
cur = l + 1 , gc = gcd( gc , a[l + 1] );
}
}

int q;
q=read();
printf("Case #%d:\n",++kcase);
while(q--){
int l=read(),r=read(),g=getST(l,r);
printf("%d %I64d\n",g,ans[g]);
}
}
}
return 0;
}