Codeforces Round #226 (Div. 2)--C. Bear and Prime Numbers
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Recently, the bear started studying data structures and faced the following problem.
You are given a sequence of integers x1, x2, ..., xn of length n and m queries, each of them is characterized by two integers li, ri. Let's introduce f(p) to represent the number of such indexes k, that xk is divisible by p. The answer to the query li, ri is the sum: , where S(li, ri) is a set of prime numbers from segment [li, ri] (both borders are included in the segment).
Help the bear cope with the problem.
The first line contains integer n (1 ≤ n ≤ 106). The second line contains n integers x1, x2, ..., xn (2 ≤ xi ≤ 107). The numbers are not necessarily distinct.
The third line contains integer m (1 ≤ m ≤ 50000). Each of the following m lines contains a pair of space-separated integers, li and ri(2 ≤ li ≤ ri ≤ 2·109) — the numbers that characterize the current query.
Print m integers — the answers to the queries on the order the queries appear in the input.
65 5 7 10 14 1532 113 124 4
970
72 3 5 7 11 4 828 102 123
07
#include<iostream>#include<cstdio>#include<vector>#include<map>#include<set>#include<algorithm>#include<cmath>#include<cctype>#include<cstdlib>#include<string>#include<cstring>#define mem(x) memset(x,0,sizeof(x));#define mem1(x) memset(x,-1,sizeof(x));using namespace std;const int maxn = 1e7+7;const int maxt = 1e6+5;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long ll;typedef unsigned long long llu;int flag[maxn],v[maxn],a[maxn];int main(){ int n; while(~scanf("%d",&n)){ mem(v); mem(a); for (int i = 0; i < n; ++i){ int k; scanf("%d",&k); a[k]++; } mem1(flag); for (int i = 2; i < maxn; ++i){ if (flag[i]){ for (int j = i; j < maxn; j+= i){ flag[j] = 0; v[i]+=a[j]; } } } for (int i =2 ; i < maxn; ++i) v[i]+=v[i-1]; int lef; scanf("%d",&lef); while(lef--) { int l,r; scanf("%d%d",&l,&r); if (l >= maxn)l=maxn-1; if (r >= maxn)r=maxn-1; printf("%d\n",v[r]-v[l-1]); } } return 0;}
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