【BZOJ2301】Problem b（莫比乌斯反演）

题意：对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，

且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

思路：第一题反演……

利用容斥原理将原询问拆成4个，问题就转化为：

1<=i<=trunc(a div k)，1<=j<=trunc(b div k)，gcd(i,j)=1的(i,j)数对个数

令f(i)表示满足gcd(x,y)=i时(x,y)的对数，F(i)表示满足i|gcd(x,y)的(x,y)的对数

显然F(i)=trunc(n div i)*trunc(m div i)

f(1)=sigma u(d)*n div d*m div d (1<=d<=n)

观察可得n div d*m div d只有2根号n个取值，且每个取值对应的u(i)是连续的一段

然后就可以记录u的前缀和来优化

From  http://m.blog.csdn.net/article/details?id=50590197

//uses sysutils;
const max=50000;
var mu,sum,prime:array[0..max]of longint;
    flag:array[0..max]of longint;
    a,b,c,d,k,i,j,t,m,cas,v:longint;
    tmp:double;

procedure swap(var x,y:longint);
var t:int64;
begin
 t:=x; x:=y; y:=t;
end;

function clac(n,m:longint):int64;
var i,t1,t2,pos:longint;
    x,y:int64;
begin
 if n>m then swap(n,m);
 clac:=0; i:=1;
 while i<=n do
 begin
  x:=n div i; y:=m div i;
  t1:=n div x;
  t2:=m div y;
  if t1<t2 then pos:=t1
   else pos:=t2;
  clac:=clac+x*y*(sum[pos]-sum[i-1]);
  i:=pos+1;
 end;
end;

begin
 assign(input,'bzoj2301.in'); reset(input);
 assign(output,'bzoj2301.out'); rewrite(output);
// tmp:=now;
 read(cas);
 mu[1]:=1;
 for i:=2 to max do
 begin
  if flag[i]=0 then
  begin
   inc(m); prime[m]:=i;
   mu[i]:=-1;
  end;
  j:=1;
  while (j<=m)and(prime[j]*i<=max) do
  begin
   t:=prime[j]*i;
   flag[t]:=1;
   if i mod prime[j]=0 then
   begin
    mu[t]:=0;
    break;
   end;
   mu[t]:=-mu[i];
   inc(j);
  end;
 end;
 for i:=1 to max do sum[i]:=sum[i-1]+mu[i];
 for v:=1 to cas do
 begin
  read(a,b,c,d,k);
  dec(a); dec(c);
  a:=a div k; b:=b div k; c:=c div k; d:=d div k;
  writeln(clac(b,d)-clac(b,c)-clac(a,d)+clac(a,c));
 end;
 //writeln((now-tmp)*86400:0:2);
 close(input);
 close(output);
end.

 UPD（2018.10.22）：C++

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define fi first
#define se second
#define MP make_pair
#define N   51000
#define M   410000
#define eps 1e-8
#define pi  acos(-1)
#define oo  1e9
 
int mu[N+10],s[N+10],prime[N+10],flag[N+10];
  
ll calc(int n,int m)
{
    if(n>m) swap(n,m);
    ll ans=0; 
    int i=1;
    while(i<=n)
    {
        ll x=n/i;
        ll y=m/i;
        int t1=n/x;
        int t2=m/y;
        int pos=min(t1,t2);
        ans+=x*y*(s[pos]-s[i-1]);
        i=pos+1;
    }
    return ans;
}
 
int main()
{
    int cas;
    scanf("%d",&cas);
    mu[1]=1;
    int m=0;
    for(int i=2;i<=N;i++)
    {
        if(!flag[i])
        {
            prime[++m]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=m;j++)
        {
            int t=prime[j]*i;
            if(t>N) break;
            flag[t]=1;
            if(i%prime[j]==0) 
            {
                mu[t]=0;
                break;
            }
            mu[t]=-mu[i];
        }
    }
    for(int i=1;i<=N;i++) s[i]=s[i-1]+mu[i];
    while(cas--)
    {
        int a,b,c,d,k;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        a--; c--;
        a/=k; b/=k; c/=k; d/=k;
        ll ans=calc(b,d)-calc(b,c)-calc(a,d)+calc(a,c);
        printf("%lld\n",ans);
    }
    return 0;
}