Sum Of Gcd

题目连接：

Description

Given you a sequence of number a1, a2, ..., an, which is a permutation of 1...n.

You need to answer some queries, each with the following format:

Give you two numbers L, R, you should calculate sum of gcd(a[i], a[j]) for every L <= i < j <= R.

Input

First line contains a number T(T <= 10),denote the number of test cases.

Then follow T test cases.

For each test cases,the first line contains a number n(1<=n<= 20000).

The second line contains n number a1,a2,...,an.

The third line contains a number Q(1<=Q<=20000) denoting the number of queries.

Then Q lines follows,each lines contains two integer L,R(1<=L<=R<=n),denote a query.

Output

For each case, first you should print "Case #x:", where x indicates the case number between 1 and T.

Then for each query print the answer in one line.

Sample Input

1

5

3 2 5 4 1

3

1 5

2 4

3 3

Sample Output

Case #1:

11

4

0

Hint

题意

给你n个数，然后Q次询问，每次问你l,r区间的两两之间的GCD和是多少

题解：

莫队+反演，直接暴力莽就好了……

代码

#include 
    
     using namespace std;const int maxn = 2e4 + 15;int unit , a[maxn] , N , M , cnt[maxn];long long ans[maxn] , phi[maxn];vector < int > factor[maxn];struct Query{    int l , r  , idx;    friend bool operator < (const Query & a , const Query & b){        int x1 = a.l / unit , x2 = b.l / unit;        if( x1 != x2 ) return x1 < x2;        return a.r < b.r;    }}Q[maxn];void Init(){    for(int i = 1 ; i < maxn ; ++ i)        for(int j = i ; j < maxn ; j += i)            factor[j].push_back( i );    phi[1] = 1;    for(int i = 2 ; i < maxn ; ++ i)        if( !phi[i] )            for(int j = i ; j < maxn ; j += i){                if( !phi[j] ) phi[j] = j;                phi[j] = phi[j] *  ( i - 1 ) / i;            }}long long add( int x ){    long long res = 0;    for( auto d : factor[x] ) res += cnt[d] * phi[d];    for( auto d : factor[x] ) cnt[d] ++ ;    return res;}long long del( int x ){    long long res = 0;    for( auto d : factor[x] ) cnt[d] -- ;    for( auto d : factor[x] ) res += cnt[d] * phi[d];    return -res;}void solve(){    memset( cnt , 0 , sizeof( cnt ) );    int l = 1 , r = 0;    long long cur = 0;    for(int i = 1 ; i <= M ; ++ i){        while( l < Q[i].l ) cur += del( a[l++] );        while( l > Q[i].l ) cur += add( a[--l] );        while( r < Q[i].r ) cur += add( a[++r] );        while( r > Q[i].r ) cur += del( a[r--] );        ans[Q[i].idx] = cur;    }}int main( int argc , char * argv[] ){    Init();    int Case , cas = 0;    scanf("%d",&Case);    while(Case--){        scanf("%d",&N);        for(int i = 1 ; i <= N ; ++ i) scanf("%d" , a + i);        scanf("%d",&M);        for(int i = 1 ; i <= M ; ++ i){            Q[i].idx = i;            scanf("%d%d",&Q[i].l,&Q[i].r);        }        unit = sqrt( N );        sort( Q + 1 , Q + M + 1 );        solve();        printf("Case #%d:\n" , ++ cas);        for(int i = 1 ; i <= M ; ++ i) printf("%lld\n" , ans[i]);    }    return 0;}