【BZOJ2301】 [HAOI2011]Problem b

problem

---

analysis

• 反演例题

• 由容斥可得，$ans=ans(b,d)-ans(a-1,d)-ans(b,c-1)+ans(a-1,c-1)$

• 单分块还不会？

• 于是每一个分块都是$O(\sqrt n+\sqrt m)$的时间，总$O(n(\sqrt n+\sqrt m))$时间复杂度

---

code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 10000001 

using namespace std;

bool bz[MAXN];
int p[MAXN],phi[MAXN],mu[MAXN],pre[MAXN];
int n,a,b,c,d,k,tot;

int read()
{
    int x=0,f=1;
    char ch=getchar();
    while (ch<'0' || '9'<ch)
    {
        if (ch=='-')f=-1;
        ch=getchar();   
    }
    while ('0'<=ch && ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

void init() 
{
    memset(bz,1,sizeof(bz));
    mu[1]=phi[1]=1,tot=0;
    for (int i=2;i<MAXN;i++) 
    {
        if (bz[i])p[tot++]=i,phi[i]=i-1,mu[i]=-1;
        for (int j=0;j<tot && i*p[j]<=MAXN;j++) 
        {
            bz[i*p[j]]=0;
            if (i%p[j]==0) 
            {
                mu[i*p[j]]=0;
                phi[i*p[j]]=phi[i]*p[j];
                break;
            } 
            mu[i*p[j]]=-mu[i],phi[i*p[j]]=phi[i]*(p[j]-1);
        }
    }
    for (int i=1;i<MAXN;i++)pre[i]=pre[i-1]+mu[i];
}

int get(int n,int m,int d) 
{
    if (n>m)swap(n,m);
    int ans=0;
    n/=d,m/=d;
    for (int i=1,last=1;i<=n;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        ans+=(pre[last]-pre[i-1])*(n/i)*(m/i);
    } 
    return ans;
}

int main()
{
    //freopen("readin.txt","r",stdin); 
    init();
    n=read();
    while (n--)
    {
        a=read(),b=read(),c=read(),d=read(),k=read();
        printf("%d\n",get(b,d,k)-get(a-1,d,k)-get(b,c-1,k)+get(a-1,c-1,k));
    }
    return 0;
}