題目連結：

題目大意：

給出一些數，每次給出一個區間，所有的數兩兩比較，如果a能整除b得一分，b能整除a得一分，問沒有得滿分的有多少個。

題目分析：

• 我們可以知道，每個區間的數最多隻有一種數字滿足條件，也就是所有數的gcd，那麼我們利用線段樹，維護某一個區間的gcd,和這個gcd出現的次數即可，無修改的線段樹水題
  

AC程式碼：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 100007
 


using namespace std;

typedef pair<int,int> PII;
int n,m;
int a[MAX];

struct Tree
{
    int l,r,gcd,num;
}tree[MAX<<2];

int gcd ( int a , int b )
{
    return !b?a:gcd(b,a%b);
}

void push_up ( int u )
{
    int a = tree[u<<1].gcd;
    int b = tree[u<<1|1].gcd;
    int x = tree[u<<1
 
].num;
    int y = tree[u<<1|1].num;
    int d = gcd ( a , b );
    tree[u].gcd = d;
    tree[u].num = 0;
    if ( d == a ) 
        tree[u].num += x;
    if ( d == b )
        tree[u].num += y;
}

void build ( int u , int l , int r )
{
    tree[u].l = l;
    tree[u].r = r;
    if ( l == r )
    {
        tree[u].gcd = a[l];
        tree[u].num = 1
 
;
        return;
    }
    int mid = l+r>>1;
    build ( u<<1 , l , mid );
    build ( u<<1|1 , mid+1 , r );
    push_up ( u );
}

PII query ( int u , int left , int right )
{
    int l = tree[u].l;
    int r = tree[u].r;
    if ( left <= l && r <= right )
        return make_pair ( tree[u].gcd , tree[u].num );
    PII ret,temp;
    ret.first = -1;
    int mid = l+r>>1;
    if ( left <= mid && right >= l ) 
        ret = query ( u<<1 , left , right );
    if ( left <= r && right > mid )
    {
        temp = query ( u<<1|1 ,left , right );
        if ( ret.first == -1 ) ret = temp;
        else
        {
            int d = gcd ( ret.first , temp.first );
            if ( d != ret.first ) ret.second = 0;
            if ( d == temp.first ) ret.second += temp.second;
            ret.first = d;
        }
    }
    return ret;
}

int main ( )
{
    while ( ~scanf ( "%d" , &n  ) )
    {
        for ( int i = 1 ; i <= n ; i++ )
            scanf ( "%d" , &a[i] );
        build ( 1 , 1 , n );
        scanf ( "%d" , &m );
        while ( m-- )
        {
            int l,r;
            scanf ( "%d%d" , &l , &r );
            printf ( "%d\n" , r-l+1-query ( 1, l , r ).second );
        }
    }
}