Sum Of Gcd
题目连接：
http://acm./showproblem.php?pid=4676
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 <bits/stdc++.h>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;}
