hdu 5726 GCD 解题报告

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 forgcd(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

151 2 4 6 741 52 43 44 4

Sample Output

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 forgcd(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

151 2 4 6 741 52 43 44 4

Sample Output

Case #1:1 82 42 46 1

题意：

给你n个数字，现在问你q次L,R区间的gcd值以及有多少个区间的gcd值和L，R相等（L，R本身也算一个）

思路：

先用rmq预处理计算任意区间的gcd值；然后计算和该区间gcd相等的区间有多少个，代码如下：

#include <bits/stdc++.h>//建议从主函数开始看using namespace std;#define LL long longconst int maxn = 100005;int dp[maxn][20];//dp[i][j]代表从第i位开始往后(1<<j)位的gcd值int num[maxn],n;int gcd(int a,int b){    return a==0?b:gcd(b%a,a);}void init(){    memset(dp,0,sizeof(dp));    for(int i=1;i<=n;i++)dp[i][0]=num[i];//（1<<0还是1嘛）    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 rmq(int l,int r){    int t=0;    while(1<<(t+1)<=(r-l+1))t++;    return gcd(dp[l][t],dp[r+1-(1<<t)][t]);//如果我们查询一段区间的gcd值的话，那么图一和图二的查询结果一定是一样的，这个可以自己证）}map<int,LL>m;//m[i]代表gcd值为i的区间有多少个int main(){    int test;    scanf("%d",&test);    int T=0;    while(test--){        T++;        m.clear();        scanf("%d",&n);        for(int i=1;i<=n;i++)scanf("%d",&num[i]);        init();        for(int i=1;i<=n;i++){//这里统计各区间的gcd值            int l=i,r=i;//当你计算l,r的区间gcd值时随着r的增加这个值一定是递减的，所以我们可以这样计算：//首先枚举i端点，然后二分找到一个最右边的r使区间(l,r)的gcd等于(l,l)，也就是说此时(l,r+1)是一个比之前小的gcd值，再如法炮制从(l,r+1)找有多少个和（l,r+1）//gcd值相等的区间，由于不知道什么鬼的证明说对于一个端点i最多能找出来log2(n)个gcd值，而每次找的复杂度为log2(n),所以这里整体复杂度为n*logn*logn;       while(r<=n){                int ll=r;                int rr=n;                int v=rmq(l,r);                while(ll<=rr){//二分找有多少个区间gcd等于(l,r)                    int mid=(rr+ll)/2;                    if(rmq(l,mid)>=v)ll=mid+1;                    else rr=mid-1;                }                m[v]+=ll-r;                r=ll;//更新r,这样可以继续找            }        }        int q;        scanf("%d",&q);        printf("Case #%d:\n",T);        while(q--){            int L,R;            scanf("%d%d",&L,&R);            int ans=rmq(L,R);            printf("%d %lld\n",ans,m[ans]);        }    }}