@牛客网暑期ACM多校训练营（第三场） H: Diff-prime Pairs ( 素数)

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链接：https://www.nowcoder.com/acm/contest/141/H
来源：牛客网
 

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.

输入描述:

Input has only one line containing a positive integer N.1 ≤ N ≤ 107

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

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3

输出

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2

示例2

输入

复制

5

输出

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6

 

[题意]:求pair(i,j) : 满足  i/gcd(i,j)  和 j/gcd(i,j)  都是素数的 个数 在 n 内

[思路]

    素数打表,  可以满足的 只有 两种形式,  1: i,j 都是素数, 2: i,j 都是素数的整数倍例如:4,6

    打完表后,二分查找最小的位置,

 

[代码]

#include <bits/stdc++.h>using namespace std;const int N = 1e7+10;
int vis[N+2],prime[N+2],cot;
typedef long long ll;
void init ()
{cot=0;for (int i = 2; i <= N; i++){if (!vis[i])prime[++cot] = i;for (int j = 1; j <= cot&& i*prime[j]<=N; j++){vis[i * prime[j]] = 1;if ((i%prime[j]) == 0) // I ÊǺÏÊý²¢ÇÒ I %prime[j]=0;break;}}
}int main()
{init();int n;ll ans = 0;scanf("%d",&n);for(int i = 1;i <= n/2;i++){ll pos = upper_bound(prime+1,prime+cot+1,n/i) - (prime +1);if(pos <= 1)break;ans += 1ll*(pos*(pos-1));}printf("%lld\n",ans);return 0;
}

 

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