GCD
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2530 Accepted Submission(s): 895
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.
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).
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=3*1e5+10;
int n,a[N],l[N],r[N];
struct node{
int l,r;
ll sum,lazy;
int gcdn;
void fun(ll v){
this->lazy+=v;
sum+=v*(r-l+1);
}
int mid(){
return (l+r)>>1;
}
}tree[4*N],tree2[4*N];
int gcd(int a,int b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
void build(int k,int l,int r)
{
tree[k].l=l;tree[k].r=r;
if(l==r)
{
tree[k].gcdn=a[l];
return;
}
int mid=(l+r)>>1;
build(2*k,l,mid);
build(2*k+1,mid+1,r);
tree[k].gcdn=gcd(tree[2*k].gcdn,tree[2*k+1].gcdn);
}
ll query(int k,int l,int r)
{
if(l<=tree[k].l&&tree[k].r<=r) return tree[k].gcdn;
else {
int mid=tree[k].mid();
int gcdk=0;
if(l<=mid) gcdk=query(2*k,l,r);
if(r>mid) gcdk=gcdk==0?query(2*k+1,l,r):gcd(gcdk,query(2*k+1,l,r));
return gcdk;
}
}
map<int,ll> tmp,tmpnext,ans;
map<int,ll>::iterator it;
void init()
{
tmp.clear();ans.clear();
for(int i=1;i<=n;i++)
{
tmpnext.clear();
tmpnext[a[i]]++;
for(it=tmp.begin();it!=tmp.end();++it)
{
int k=gcd(a[i],it->first);
tmpnext[k]+=it->second;
}
for(it=tmpnext.begin();it!=tmpnext.end();it++)
ans[it->first]+=it->second;
tmp=tmpnext;
}
}
int main()
{
int cas,q,kk=0;
scanf("%d",&cas);
while(cas--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
build(1,1,n);
init();
scanf("%d",&q);
for(int i=1;i<=q;i++) scanf("%d%d",&l[i],&r[i]);
printf("Case #%d:
",++kk);
for(int i=1;i<=q;i++)
{
int k=query(1,l[i],r[i]);
printf("%d %lld
",k,ans[k]);
}
}
return 0;
}
我们注意观察gcd(a_{l},a_{l+1},...,a_{r})gcd(al,al+1,...,ar)，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(al,al+1,...,ar)是递减的，同时gcd(a_{l},a_{l+1},...,a_{r})gcd(al,al+1,...,ar)最多 有log 1000,000,000log 1000,000,000个不同的值，为什么呢？因为a_{l}al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(al,al+1,...,ar)
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(al,al+1,...,ar)
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(al,al+1,...,ar)为多少，然后再在map 里面查找对应答案即可.
错因分析：1.没能正确发现对于一个a（<=1e9）他的质因数个数是loga级别的，因为比如2*2*3*5.....=a，则左边至多loga个元素，