Codeforces Round #226 (Div. 2) C. Bear and Prime Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/u011639256/article/details/18751935

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: $\text{[math]}$ , 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.

Input

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.

Output

Print m integers — the answers to the queries on the order the queries appear in the input.

       Sample test(s) 
     

         input 
       
6
5 5 7 10 14 15
3
2 11
3 12
4 4

         output 
       
9
7
0

         input 
       
7
2 3 5 7 11 4 8
2
8 10
2 123

         output 
       
0
7

Note

Consider the first sample. Overall, the first sample has 3 queries.

The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9.
The second query comes l = 3, r = 12. You need to count f(3) + f(5) + f(7) + f(11) = 1 + 4 + 2 + 0 = 7.
The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0.

英语跟不上就是不行啊。。。

红体字读了n多遍一直没理解什么意思，最后醍醐灌顶、大彻大悟

题目的意思是给出n

给出x1...xn

再给出m

给出m行l[i],r[i]

对应这m行，每行输出l[i]到r[i]（包括l[i],r[i]）所有质数的f（）函数对应值

之前一直不理解f()函数是怎么计算的

后来想通了

例如案例1中的f(2)=2

是因为x[]数组中有两个数可以被2整除

虽然思路没错，但实现起来就比较恶心了，要注意到l[i],r[i]的取值范围有多大。。。

我刚开始写了两个函数，一个是用来判断是否为质数，另一个就是f()函数

交上去第四个案例超时

又改进了一下，使得不用专门判断是否为质数，而是在f()函数中判断若不为质数让其对应f()为0即可

交上去还是第四个超时

又改进了一下，先把每个f()值保存在数组中，免得每次在循环中都要多次调用函数

交上去还是跪了。。。。

先把代码贴上吧

明天再改正：

#include <iostream>
#include <string>
#include <vector>
#define rep(i,j,k) for(int i=(j); i<k; i++)
#define maxn 100100
#define maxm 50010
#define max 20000000
int a[maxn]; 
int l[maxm],r[maxm];
long long int res[maxm];
long long b[max];
using namespace std;

int n,m;

int f(int t){
    int times = 0;
    if(a[1]%t == 0)
     times++;
    for(int i=2; i<=n; ++i){
     if(t>i && t%i==0)
      return 0;
     if(a[i]%t == 0)
      times++;
    }
    return times;
}

int main(void){
    while(cin >> n){
     rep(i,1,n+1)
      cin >> a[i];
     
     cin >> m;
     rep(i,0,m)
      cin >> l[i] >> r[i];
     for(int i=2; i<max; ++i)
      b[i]=f(i);
     for(int i=0; i<m; i++){
      for(int j=l[i]; j<=r[i]; ++j)
        res[i] += b[j];
      cout << res[i] << endl;
     }
    }
    return 0;
}

贴上正确代码以及我的理解：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

const int kMaxX = 10000000, kMaxP = 664579;
int n, pr[kMaxP + 1];
bool not_pr[kMaxX + 1];//非素数 
int sum[kMaxX + 1];

void init(){
    for(int i=2; i<=kMaxX; ++i){
     if(!not_pr[i]) pr[++pr[0]] = i;
     //找出所有的素数，存入pr数组中，pr[0]表示当前素数个数 
     for(int j=1,k; j<=pr[0] && (k=pr[j]*i)<=kMaxX; ++ j){
      not_pr[k] = true;
      //每个素数的倍数都是非素数，这样可以省去很多判断的时间 
      if(i%pr[j]==0) break;
     }
    }
}

int main(void){
    init();
    scanf("%d", &n);
    for(int x; n--; ){
     scanf("%d", &x);
     for(int i=1; pr[i]*pr[i]<= x; ++i)//用pr[i]*pr[i]<=n降低时间复杂度 
      if(x%pr[i]==0){
       ++sum[pr[i]];
       while(x%pr[i]==0) 
        x /= pr[i];
        //意在把x变小 
      }
     if(x != 1) 
     //例如x刚开始等于14，在上面的for循环中
     //只能找到14%2==0，不能找到14%7==0，使用while循环加判断则可以做到 
      ++sum[x];
    }
    for(int i=3; i<=kMaxX; ++ i) 
     sum[i]+=sum[i-1];
    //sum[i]对应r[i]的f()函数加权和 
    scanf("%d", &n);
    for(int l,r; n--; ){
     scanf("%d%d", &l, &r);
     l = min(l, kMaxX);//为什么？？？ 
     r = min(r, kMaxX);//因为对x[]数组，最大只有10,000,000
     //当l r大于kMaxX时不可能存在x满足条件 
     printf("%d\n", sum[r]-sum[l-1]);
    }
    return 0;
}