BZOJ 2005 2005: [Noi2010]能量采集 | 容斥原理

题目:

http://www.lydsy.com/JudgeOnline/problem.php?id=2005

题解:

https://www.geek-share.com/detail/2623817483.html

#include<cstdio>
#include<algorithm>
#define N 100005
using namespace std;
typedef long long ll;
ll phi
,su
,sum
;
bool he
;
void Euler(int n)
{
int tot=0;
phi[1]=1;
for(int i=2;i<=n;i++)
{
if(!he[i])
{
su[++tot]=i;
phi[i]=i-1;
}
for(int j=1;j<=tot;j++)
{
if(i*su[j]>=n)break;
he[i*su[j]]=1;
if(i%su[j]==0)
{
phi[i*su[j]]=phi[i]*su[j];break;
}
else phi[i*su[j]]=phi[i]*(su[j]-1);
}
}
for(int i=1;i<=n;i++)
phi[i]+=phi[i-1];
return;
}
int main(){
ll n,m,ans=0;
scanf("%lld%lld",&n,&m);
if(n>m)swap(n,m);
Euler(n+1);
for(ll i=1,j;i<=n;i=j+1)
{
j=min(n/(n/i),m/(m/i));
ans+=(ll)(phi[j]-phi[i-1])*(n/i)*(m/i);
}
printf("%lld\n",2*ans-n*m);
return 0;
}