BZOJ 2005 2005: [Noi2010]能量采集 | 容斥原理

题目:

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

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题解:

http://blog.csdn.net/popoqqq/article/details/39924877

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