bzoj2301

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;}