链接:https://www.nowcoder.com/acm/contest/141/H
来源:牛客网
Diff-prime Pairs
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
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.
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
3
输出
2
示例2
输入
5
输出
6
题意:给一个n,在【1,n】中找出所有的对数(i,j)( i != j) && ( i,j<=n )
并且 = gcd(i,j)为i,j 的 最大公约数
思路 : 1.先在【1,n】中找出所有的素数的个数 ans,并标记号(素数筛选法)
2. 总数 Ans=ans*(ans-1); 每 一个素数与(其他的素数)进行两两配对
比如当n=5 时 一开始 素数(2.3.5)对有Ans=ans*(ans-1)=6; {(2.3)(2.5)(3.2)(3.5)(5.2)(5.3) }
3.找出 符合条件素数对的倍数
比如当n=6时一开始 素数(2.3.5)对有Ans=ans*(ans-1)=6 还有两个符合条件的 (2*2.3*2)即(4.6)和(6.4);
代码实现:
#include<bits/stdc++.h>
using namespace std;
const int M=10000010;
bool a[M];
int sum[M];
int main()
{
int i,j,n,m;
long long ans=0;
scanf("%d",&n);
memset(a,1,sizeof a);//素数筛选法
a[1]=0;
for(i=2; i<=n; i++)
if(a[i])
{
for(j=2;j*i<=n;j++)
a[j*i]=0;
}
for(i=2;i<=n;i++)//记录一共有多少个素数
if(a[i]) ans++,sum[i]=ans;//sum[i] i点之前有一共多少个素数
ans=ans*(ans-1);//素数对
for(i=n;i>=3;i--)
{
if(a[i])//素数
{
int t=n/i;//有多少个可满足条件的 素数倍数
t--;sum[i]--;//减去素数本身
ans+=(2*t*sum[i]);//新增的非素数对 *2是因为(4.6)(6.4);
}
}printf("%lld\n",ans);
}