【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=;

 var mu,sum,prime:array[..max]of longint;

     flag:array[..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:=; i:=;

  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-]);

   i:=pos+;

  end;

 end;

 begin

  assign(input,'bzoj2301.in'); reset(input);

  assign(output,'bzoj2301.out'); rewrite(output);

 // tmp:=now;

  read(cas);

  mu[]:=;

  for i:= to max do

  begin

   if flag[i]= then

   begin

    inc(m); prime[m]:=i;

    mu[i]:=-;

   end;

   j:=;

   while (j<=m)and(prime[j]*i<=max) do

   begin

    t:=prime[j]*i;

    flag[t]:=;

    if i mod prime[j]= then

    begin

     mu[t]:=;

     break;

    end;

    mu[t]:=-mu[i];

    inc(j);

   end;

  end;

  for i:= to max do sum[i]:=sum[i-]+mu[i];

  for v:= 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)*::);

  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+],s[N+],prime[N+],flag[N+];

 ll calc(int n,int m)

 {

     if(n>m) swap(n,m);

     ll ans=;

     int i=;

     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-]);

         i=pos+;

     }

     return ans;

 }

 int main()

 {

     int cas;

     scanf("%d",&cas);

     mu[]=;

     int m=;

     for(int i=;i<=N;i++)

     {

         if(!flag[i])

         {

             prime[++m]=i;

             mu[i]=-;

         }

         for(int j=;j<=m;j++)

         {

             int t=prime[j]*i;

             if(t>N) break;

             flag[t]=;

             if(i%prime[j]==)

             {

                 mu[t]=;

                 break;

             }

             mu[t]=-mu[i];

         }

     }

     for(int i=;i<=N;i++) s[i]=s[i-]+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 ;

 }

巴特西

【BZOJ2301】Problem b（莫比乌斯反演）

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