GCD
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 5030 Accepted Submission(s): 1800
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.
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).
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
/**
题意:给定一个长度为n的序列a
l r 表示[l,r]的连续区间
问:gcd(l,l+1,l+2.....r)=ans;
a序列中存在多少个连续区间使得该连续区间的gcd等于ans;
解题思路:对于一段连续区间的gcd 可由RMQ直接得到
由于求解的是子区间的数量 然而对于连续区间子序列gcd值不同数的个数 为lgn级别的 并且这些数是发散的 (1e9的数据)
因此可以采用最暴力的方法map进行存储答案 最后O(1)查询;
中间过程状态 : 固定左端点 枚举右端点 看在此gcd下的区间长度能达到的最大长度 并实时用map进行记录 二分性质为:最小值最大化;
动态更新gcd的过程就是类似于尺取法的过程
***tricks****
ST表的处理注意角标的变化;
时间复杂度 n*lgn*lgn
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e6+7;
int n,dp[maxn][20];
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
void init(){
for(int j=1;(1<<j)<=n;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 query(int l,int r){
int k=(int)(log10(r*1.0-l*1.0+1)/log10(2.0));
return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}
map<int,ll>mp;
void solve(){
mp.clear();
for(int i=1;i<=n;i++){
int tmp=dp[i][0],j=i;
while(j<=n){
int l=j,r=n;
while(l<r){
int mid=l+r+1>>1;
if(query(i,mid)==tmp) l=mid;
else r=mid-1;
}
mp[tmp]+=(l-j+1);
j=l+1;
tmp=query(i,j);
}
}
}
void solved(){
int t;scanf("%d",&t);
for(int cas=1;cas<=t;cas++){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&dp[i][0]);
init();
solve();
printf("Case #%d:\n",cas);
int q;scanf("%d",&q);
while(q--){
int l,r;scanf("%d %d",&l,&r);
int ret=query(l,r);
printf("%d %lld\n",ret,mp[ret]);
}
}
}
int main (){
solved();
return 0;
}