学步园

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2190

题解：让你求0<=x,y<n中，有多少对x,y满足gcd(x,y)=1。欧拉函数前n-1项的和*2-1，又因为有0，所以还要加上2.

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
using namespace std;

#define si1(a) scanf("%d",&a)
#define si2(a,b) scanf("%d%d",&a,&b)
#define sd1(a) scanf("%lf",&a)
#define sd2(a,b) scanf("%lf%lf",&a,&b)
#define ss1(s)  scanf("%s",s)
#define pi1(a)    printf("%d\n",a)
#define pi2(a,b)  printf("%d %d\n",a,b)
#define mset(a,b)   memset(a,b,sizeof(a))
#define forb(i,a,b)   for(int i=a;i<b;i++)
#define ford(i,a,b)   for(int i=a;i<=b;i++)

typedef long long LL;
const int N=40005;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-7;


LL phi[N],con[N];

void phi_xiaohao()
{
    LL n=40000;
    mset(phi,0);
    phi[1]=1;
    for(LL i=2;i<=n;i++)
        if(!phi[i])
        {
            for(LL j=i;j<=n;j+=i)
            {
                if(!phi[j]) phi[j]=j;
                phi[j]=phi[j]/i*(i-1);
            }
        }
    mset(con,0);
    for(LL i=1;i<=n;i++)
        con[i]=con[i-1]+phi[i];
}

int main()
{

    phi_xiaohao();

    int n;
    while(cin>>n)
    {
        cout<<2*con[n-1]-1+2<<endl;
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
using namespace std;

#define si1(a) scanf("%d",&a)
#define si2(a,b) scanf("%d%d",&a,&b)
#define sd1(a) scanf("%lf",&a)
#define sd2(a,b) scanf("%lf%lf",&a,&b)
#define ss1(s)  scanf("%s",s)
#define pi1(a)    printf("%d\n",a)
#define pi2(a,b)  printf("%d %d\n",a,b)
#define mset(a,b)   memset(a,b,sizeof(a))
#define forb(i,a,b)   for(int i=a;i<b;i++)
#define ford(i,a,b)   for(int i=a;i<=b;i++)

typedef long long LL;
const int N=40000;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-7;

LL x;
LL A,B,C,D,K;
LL mu[N+10],sum[N+10],prime[N+10];
bool com[N+1];

void GetPrimes()
{
    memset(com,0,sizeof(com));
    mu[1]=1;
    x=0;
    for(LL i=2;i<=N;++i)
    {
        if (!com[i])    { prime[x++] = i; mu[i] = -1; }
        for (LL j=0;j<x&&i*prime[j]<=N;++j)
        {
            com[i*prime[j]] = true;
            if (i%prime[j]) mu[i*prime[j]] = -mu[i];
            else { mu[i*prime[j]] = 0; break; }
        }
    }
    for (LL i=1;i<=N;++i)
        sum[i] = sum[i-1] + mu[i];
}

int main()
{
//    freopen("input.txt","r",stdin);
    GetPrimes();
    LL n;
    while(cin>>n)
    {
        n--;
        LL sum=0;
        for(LL i=1;i<=n;i++)
            sum+=mu[i]*(n/i)*(n/i);
        printf("%lld\n",sum+2);
    }
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
using namespace std;
 
#define si1(a) scanf("%d",&a)
#define si2(a,b) scanf("%d%d",&a,&b)
#define sd1(a) scanf("%lf",&a)
#define sd2(a,b) scanf("%lf%lf",&a,&b)
#define ss1(s)  scanf("%s",s)
#define pi1(a)    printf("%d\n",a)
#define pi2(a,b)  printf("%d %d\n",a,b)
#define mset(a,b)   memset(a,b,sizeof(a))
#define forb(i,a,b)   for(int i=a;i<b;i++)
#define ford(i,a,b)   for(int i=a;i<=b;i++)
 
typedef long long LL;
const int N=40000;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-7;
 
LL x;
LL A,B,C,D,K;
LL mu[N+10],sum[N+10],prime[N+10];
bool com[N+1];
 
void GetPrimes()
{
    memset(com,0,sizeof(com));
    mu[1]=1;
    x=0;
    for(LL i=2;i<=N;++i)
    {
        if (!com[i])    { prime[x++] = i; mu[i] = -1; }
        for (LL j=0;j<x&&i*prime[j]<=N;++j)
        {
            com[i*prime[j]] = true;
            if (i%prime[j]) mu[i*prime[j]] = -mu[i];
            else { mu[i*prime[j]] = 0; break; }
        }
    }
    for (LL i=1;i<=N;++i)
        sum[i] = sum[i-1] + mu[i];
}
 
LL Process(LL n,LL m)//gcd(x,y)=1的对数,1<=x<=n,1<=y<=m
{
    LL res=0;
    if(n>m) swap(n,m);
    for(LL i=1,last=0;i<=n;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        res+=(sum[last]-sum[i-1])*(n/i)*(m/i);
    }
    return res;
}
 
int main()
{
//    freopen("input.txt","r",stdin);
    GetPrimes();
    LL n;
    while(cin>>n)
    {
 
        LL ans=0;
        ans+=Process(n-1,n-1)+2;
        printf("%lld\n",ans);
    }
}