CF385C Bear and Prime Numbers 数学-CFANZ编程社区

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 CF385C Bear and Prime Numbers 数学_#define

Help the bear cope with the problem.

输入输出格式

输入格式：

The first line contains integer CF385C Bear and Prime Numbers 数学_#define_02

The third line contains integer CF385C Bear and Prime Numbers 数学_#define_03

输出格式：

Print CF385C Bear and Prime Numbers 数学_i++_04

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

9
7
0

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
}

EPFL - Fighting