jzoj 3623【SDOI2014】数表(table)

Description

Input

Output

Sample Input

Sample Output

20
148

Data Constraint

解题思路

Source Code

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;

const int N=100000;

struct note
{
int n,m,a,ans,da;
};

struct note2
{
int x,y;
};

note s
;
note2 a
;
int g
,f
,mu
,t
,prime
;
int q,maxn;

bool comp1(note x,note y)
{
return x.a<y.a;
}

bool comp2(note x,note y)
{
return x.da<y.da;
}

bool comp3(note2 x,note2 y)
{
return x.x<y.x;
}

void find()
{
memset(t,0,sizeof(t));
mu[1]=1;
for (int i=2;i<=maxn;++i)
{
if (t[i]==0)
{
prime[++prime[0]]=i;
mu[i]=-1;
}
for (int j=1;j<=prime[0];++j)
{
int tt=i*prime[j];
if (tt>maxn) break;
t[tt]=1;
if (i%prime[j]==0) {mu[tt]=0; break; }
mu[tt]=-mu[i];
}
}
memset(f,0,sizeof(f));
int sq=floor(sqrt(maxn));
for (int i=1;i<=sq;++i)
for (int j=i;j<=maxn;j+=i)
{
f[j]+=i;
if ((j/i)>sq) f[j]+=j/i;
}
}

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

void init()
{
q=read();
for (int i=1;i<=q;++i)
{
s[i].n=read();
s[i].m=read();
s[i].a=read();
s[i].da=i; s[i].ans=0;
maxn=max(maxn,min(s[i].n,s[i].m));
}
find();
//for (int i=1;i<=maxn;++i) printf("%d ",mu[i]); printf("\n");
//for (int i=1;i<=maxn;++i) printf("%d ",f[i]);  printf("\n");
for (int i=1;i<=maxn;++i)
{
a[i].x=f[i];
a[i].y=i;
}
sort(s+1,s+q+1,comp1);
sort(a+1,a+maxn+1,comp3);
//for (int i=1;i<=maxn;++i) printf("%d %d\n",a[i].x,a[i].y);
}

void update(int x,int value)
{
for (int i=x;i<=maxn;i+=i&(-i))
g[i]+=value;
}

void insert(int d)
{
for (int i=1;i<=maxn/d;++i)
if (mu[i]!=0) update(i*d,f[d]*mu[i]);
}

int getsum(int x)
{
int ans=0;
for (int i=x;i>0;i-=i&(-i))
ans+=g[i];
return ans;
}

int main()
{
freopen("table.in","r",stdin);
freopen("table.out","w",stdout);
init();
int l=1;
for (int k=1;k<=q;++k)
{
while (a[l].x<=s[k].a && l<=maxn)
{
insert(a[l].y);
++l;
}
//for (int i=1;i<=maxn;++i) printf("%d ",getsum(i)-getsum(i-1)); printf("\n");
int i,p,r;
i=1;
int n=min(s[k].n,s[k].m);
int m=max(s[k].n,s[k].m);
while (i<=n)
{
p=min(n/(n/i),m/(m/i));
r=getsum(p)-getsum(i-1);
s[s[k].da].ans+=r*(n/i)*(m/i);
i=p+1;
}
}
for (int i=1;i<=q;++i)
printf("%d\n",s[i].ans&0x7fffffff);
fclose(stdin); fclose(stdout);
}

PS

刚开始写的不够优美，只有70分，后来把lowbit（x）这个专门的函数去掉了，就过了。。。┭┮﹏┭┮