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

输入样例#1：
复制

6

5 5 7 10 14 15

3

2 11

3 12

4 4

输出样例#1： 复制

输入样例#2： 复制

7

2 3 5 7 11 4 8

2

8 10

2 123

输出样例#2： 复制

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;

}

巴特西

CF385C Bear and Prime Numbers 数学

题意翻译

题目描述

输入输出格式

输入输出样例

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