GCD Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others)
Total Submission(s): Accepted Submission(s): Problem Description
Give you a sequence of N(N≤,) integers : a1,...,an(<ai≤,,). There are Q(Q≤,) queries. For each query l,r you have to calculate gcd(al,,al+,...,ar) and count the number of pairs(l′,r′)(≤l<r≤N)such that gcd(al′,al′+,...,ar′) equal gcd(al,al+,...,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(<ai≤,,). 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 ). For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+,...,ar′) equal gcd(al,al+,...,ar). Sample Input Sample Output
Case #:
/*
难点在于计数，关键要发现gcd的递减性
对于1到n的(l,r)
对于固定的l
gcd(l,r)>=gcd(l,r+1)
对于固定的r
gcd(l,r)<=gcd(l+1,r)
因此可以不用逐一地进行计数
方法为：
枚举每一个左区间,对满足gcd(l,ri)的右区间进行二分查找
跳跃着进行计数
*/
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <map>
#include <vector>
#define scan1(x) scanf("%d",&x)
#define scan2(x,y) scanf("%d%d",&x,&y)
#define scan3(x,y,z) scanf("%d%d%d",&x,&y,&z)
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const int Max=1e6+;
int dp[Max][];
map<int,LL> vis;
int A[Max];
int gcd(int x,int y)
{
if(x<y) swap(x,y);
return (y==?x:gcd(y,x%y));
}
void RMQ_init(int n)
{
for(int i=; i<=n; i++) dp[i][]=A[i];
for(int j=; (<<j)<=n; j++)
{
for(int i=; i+(<<j)-<=n; i++)
{
dp[i][j]=gcd(dp[i][j-],dp[i+(<<(j-))][j-]);
}
}
}
int RMQ(int l,int r)
{
int k=;
while((<<(k+))<=r-l+) k++;
return gcd(dp[l][k],dp[r-(<<k)+][k]);
}
void Init()
{
vis.clear();
}
int L[Max],R[Max];
int main()
{
int T,ca=;
for(scan1(T); T; T--)
{
int n,m,num;
scan1(n);
for(int i=; i<=n; i++) scan1(A[i]);
RMQ_init(n);
Init();
scan1(m);
for(int i=; i<=m; i++)
{
scan2(L[i],R[i]);
num=RMQ(L[i],R[i]);
vis.insert(make_pair(num,));
}
int l,r,mid,d1,d2,ans,nex;
for(int i=; i<=n; i++)
{
nex=i;
while(nex<=n)
{
d1=RMQ(i,nex);
l=nex;r=n;
while(l<=r)
{
mid=(l+r)>>;
d2=RMQ(i,mid);
if(d2>=d1) l=mid+,ans=mid;
else r=mid-;
}
if(vis.find(d1)!=vis.end())
vis[d1]+=(ans-nex)+;
nex=r+;
}
}
printf("Case #%d:\n",ca++);
for(int i=; i<=m; i++)
{
ans=RMQ(L[i],R[i]);
printf("%d %lld\n",ans,vis[ans]);
}
}
return ;
}