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新视野OJ 2190 [SDOI2008]仪仗队 (数论-gcd)

2013-10-06 10:37 337 查看
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:

2190Accepted1896 kb40 msC++/Edit1517 B
#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
,con
;

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);
}
}
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