2018牛客网暑期ACM多校训练营(第三场) H - Diff-prime Pairs - [欧拉筛法求素数]
题目链接:https://www.nowcoder.com/acm/contest/141/H
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 262144K,其他语言524288K
64bit IO Format: %lld
空间限制:C/C++ 262144K,其他语言524288K
64bit IO Format: %lld
题目描述
Eddy has solved lots of problem involving calculating the number of coprime pairs within some range. This problem can be solved with inclusion-exclusion method. Eddy has implemented it lots of times. Someday, when he encounters another coprime pairs problem, he comes up with diff-prime pairs problem. diff-prime pairs problem is that given N, you need to find the number of pairs (i, j), where
and
are both prime and i ,j ≤ N. gcd(i, j) is the greatest common divisor of i and j. Prime is an integer greater than 1 and has only 2 positive divisors.
Eddy tried to solve it with inclusion-exclusion method but failed. Please help Eddy to solve this problem.
Note that pair (i1, j1) and pair (i2, j2) are considered different if i1 ≠ i2 or j1 ≠ j2.
Eddy tried to solve it with inclusion-exclusion method but failed. Please help Eddy to solve this problem.
Note that pair (i1, j1) and pair (i2, j2) are considered different if i1 ≠ i2 or j1 ≠ j2.
输入描述:
Input has only one line containing a positive integer N.
1 ≤ N ≤ 10^7
输出描述:
Output one line containing a non-negative integer indicating the number of diff-prime pairs (i,j) where i, j ≤ N
示例1
输入
3
输出
2
示例2
输入
5
输出
6
题意:
给出一个数字 n (1 ≤ n ≤ 1e7),求多少 数对(i, j) 满足 $\frac{i}{{\gcd \left( {i,j} \right)}}$ 和 $\frac{j}{{\gcd \left( {i,j} \right)}}$ 均为质数,且1 ≤ i, j ≤ n。
题解:
筛出[1,n]之间所有的素数,
不难知道,每次取到其中两个素数组成一个素数对(x, y),不妨设 x < y,那么相应的就增加了 $2 \times \left\lfloor {n/y} \right\rfloor $ 个数对;
例如,n=7,取到素数对(2,3),那么 $\left\lfloor {n/3} \right\rfloor = \left\lfloor {7/3} \right\rfloor = 2$,就有 $2 \times 2 = 4$ 个数对:(1*2,1*3) = (2,3)、(3,2)、(2*2,2*3) = (4,6)、(6,4);
对欧拉筛法稍加改造,添加一行代码即可。时间复杂度O(n)。
AC代码:
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e7+5; int n; ll ans; bool isPrime[maxn]; int prime[maxn/10],cnt; void screen()//欧拉筛法求素数 { cnt=0; memset(isPrime,1,sizeof(isPrime)); isPrime[0]=isPrime[1]=0; for(int i=2;i<=n;i++) { if(isPrime[i]) { prime[cnt++]=i; ans+=2*(n/i)*(cnt-1); //每找到一个素数i,其就可以与前面所有出现过的cnt-1个素数组成cnt-1个素数对,相应的就有2*(n/i)*(cnt-1)个数对 } for(int j=0;j<cnt;j++) { if(i*prime[j]>n) break; isPrime[(i*prime[j])]=0; if(i%prime[j]==0) break; } } } int main() { scanf("%d",&n); ans=0; screen(); cout<<ans<<endl; }
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