bzoj2301 - 莫比乌斯反演

bzoj2301
莫比乌斯反演
分块处理
参考了PoPoQQQ神的ppt

不懂为什么读入用cin就RE。QAQ

#include <bits/stdc++.h>

using namespace std;

const int N=50000+10;

typedef long long LL;
int prime[N],tot;
int mu[N];
int sum[N];
void Mu(){
    mu[1]=1;
    for (int i=2;i<N;i++){
        if (!prime[i]){
            prime[++tot] = i;
            mu[i]=-1;
        }
        for (int j=1;prime[j]*i<N;j++){
            prime[ prime[j]*i ] = 1;
            if (i%prime[j]==0){
                mu[i*prime[j]] = 0;
                break;
            }
            mu[i*prime[j]] = -mu[i];
        }
    }
    for (int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
}
LL calc(int a,int b){
    if (a>b)swap(a,b);
    int last=0;
    LL ans=0;
    for (int i=1;i<=a;i=last+1){
        last=min(a/(a/i),b/(b/i));
        ans+=(LL)(a/i)*(b/i)*(sum[last]-sum[i-1]);
    }
    return ans;
}
void work(){
    int a,b,c,d,K;
    //cin>>a>>b>>c>>d>>K;
    scanf("%d%d%d%d%d",&a,&b,&c,&d,&K);
    LL ans = calc(b/K,d/K)-calc(b/K,(c-1)/K)-calc((a-1)/K,d/K)+calc((a-1)/K,(c-1)/K);
    printf("%lld\n",ans);
    //
    //cout<<calc(b/K,d/K)-calc(b/K,(c-1)/K)-calc((a-1)/K,d/K)+calc((a-1)/K,(c-1)/K)<<endl;
    //cout<<calc(b/K,d/K)<<" "<<calc(b/K,(c-1)/K)<<" "<<calc((a-1)/K,d/K)<<" "<<calc((a-1)/K,(c-1)/K)<<endl;
}
int main()
{
    //freopen("1.txt","r",stdin);
    Mu();
    int Case;cin>>Case;
    while (Case--){
        work();
    }
    return 0;
}