CF385C Bear and Prime Numbers 数学
题意翻译
给你一串数列a.对于一个质数p,定义函数f(p)=a数列中能被p整除的数的个数.给出m组询问l,r,询问[l,r]区间内所有素数p的f(p)之和.
题目描述
Recently, the bear started studying data structures and faced the following problem.
You are given a sequence of integers x1,x2,...,xn x_{1},x_{2},...,x_{n} x1,x2,...,xn of length n n n and m m m queries, each of them is characterized by two integers li,ri l_{i},r_{i} li,ri . Let's introduce f(p) f(p) f(p) to represent the number of such indexes k k k , that xk x_{k} xk is divisible by p p p . The answer to the query li,ri l_{i},r_{i} li,ri is the sum: , where S(li,ri) S(l_{i},r_{i}) S(li,ri) is a set of prime numbers from segment [li,ri] [l_{i},r_{i}] [li,ri] (both borders are included in the segment).
Help the bear cope with the problem.
输入输出格式
输入格式:
The first line contains integer n n n (1<=n<=106) (1<=n<=10^{6}) (1<=n<=106) . The second line contains n n n integers x1,x2,...,xn x_{1},x_{2},...,x_{n} x1,x2,...,xn (2<=xi<=107) (2<=x_{i}<=10^{7}) (2<=xi<=107) . The numbers are not necessarily distinct.
The third line contains integer m m m (1<=m<=50000) (1<=m<=50000) (1<=m<=50000) . Each of the following m m m lines contains a pair of space-separated integers, li l_{i} li and ri r_{i} ri (2<=li<=ri<=2⋅109) (2<=l_{i}<=r_{i}<=2·10^{9}) (2<=li<=ri<=2⋅109) — the numbers that characterize the current query.
输出格式:
Print m m m integers — the answers to the queries on the order the queries appear in the input.
输入输出样例
7 2 3 5 7 11 4 8 2 8 10 2 123
0 7
首先可以线性筛筛出 1e7 里面所有的素数;
对于每一个x[i] ,我们记录其次数,然后类似于埃筛的做法,对于每个素数,用 sum [ i ] 来累计有该素数因子的数的个数;
最后用前缀和维护;
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize("O3") using namespace std; #define maxn 10000005 #define inf 0x3f3f3f3f #define INF 9999999999 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-3 typedef pair<int, int> pii; #define pi acos(-1.0) const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } ll sqr(ll x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans; } int n; int x[maxn]; int m; int cnt[maxn]; int prime[maxn]; int tot = 0; int sum[maxn]; bool vis[maxn]; void init() { vis[1] = 1; for (int i = 2; i < maxn; i++) { if (!vis[i])prime[++tot] = i; for (int j = 1; prime[j] * i < maxn; j++) { vis[prime[j] * i] = 1; if (i%prime[j] == 0)break; } } } int main() { //ios::sync_with_stdio(0); rdint(n); for (int i = 1; i <= n; i++)rdint(x[i]), cnt[x[i]]++; init(); for (int i = 1; i <= tot; i++) { for (int j = 1; j*prime[i] < maxn; j++) { sum[i] += cnt[j*prime[i]]; } } for (int i = 1; i <= tot; i++)sum[i] += sum[i - 1]; rdint(m); while (m--) { int l, r; rdint(l); rdint(r); int pos1 = upper_bound(prime + 1, prime + 1 + tot, r) - prime - 1; int pos2 = lower_bound(prime + 1, prime + 1 + tot, l) - prime - 1; // cout << pos1 << ' ' << pos2 << endl; cout << sum[pos1] - sum[pos2] << endl; } return 0; }