牛客網暑期ACM多校訓練營（第三場） H Diff-prime Pairs

題目鏈接
 

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.

輸入描述:

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

輸出描述:

Output one line containing a non-negative integer indicating the number of diff-prime pairs (i,j) where i, j ≤ N

示例1

輸入

3

輸出

2

示例2

輸入

5

輸出

6

 

題意：給你一個n,讓你求從1到n中滿足 i/gcd(i,j)和j/gcd(i,j) 都屬素數的（i，j）的個數。

解題思路：

如果i，j自己就是素數的話，那麼肯定滿足條件，那麼如果我們一開始就知道從1到n的範圍裏素數的個數的話，那麼我們就可以得到答案的一部分解，爲什麼說是一部分，因爲還有一部分解來自於素數的倍數（>1）。我們可以知道如果在1~n中，能被i整除的數的個數爲n/i 素數已經用完了，那麼我們把這個答案-1爲剩下的倍數。

然後我只需要統計一下兩個 素數的倍數的組合方案數就ok了。我一開始暴力跑，但是超時了，借用了隊友的思想。非常棒。

但是竟然還多次提交浮點錯誤，真是醉了。還是太粗心。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e7+10;
#define ll long long
ll cnt[maxn],tot,pri[maxn],n;
bool vis[maxn];
void find(){
	memset(vis,0,sizeof(vis));tot=0;
	vis[1]=1;
	ll i,j;
	for(i=2;i<=1e7;i++){
		if(!vis[i]) pri[tot++]=i;
		for(j=0;j<tot&&i*pri[j]<=1e7;j++){
			vis[i*pri[j]]=1;
		} 
	}
}
int main(){
	ll i,j;
	find();
	scanf("%lld",&n);
	ll l;
	for(l=0;pri[l]<=n&&l<tot;l++);
	ll ans=l*(l-1)/2;
	for(i=1;i<l;i++){
		cnt[i]=n/pri[i]-1;
		ans+=i*cnt[i]; 
	}
	printf("%lld\n",ans*2);
	return 0;
}