新视野OJ 2190 [SDOI2008]仪仗队 (数论-gcd)
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:
2190 | Accepted | 1896kb | 40ms | C++/Edit | 1517 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[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;
}