spoj Primes in GCD Table 莫比乌斯反演

Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has created a GCD (greatest common divisor) table! So he now has a table (of height a and width b), indexed from (1,1) to (a,b), and with the value of field (i,j) equal to gcd(i,j). He wants to know how many times he has used prime numbers when writing the table.

Input

First, t ≤ 10, the number of test cases. Each test case consists of two integers, 1 ≤a,b < 10⁷.

Output

For each test case write one number - the number of prime numbers Johnny wrote in that test case.

Example

Input:210 10100 100Output:302791

莫比乌斯的理解见  传送门

题意：

要求求出gcd（x , y）为质数的个数，x ，y的区间见题

题解：

                  令        f(p)=gcd(x,y)为p       (p为质数)

                     F(pk)=gcd(x,y)>=p   (k为质数)

公式借鉴    传送门

ans=∑pmin(n,m)f(p)=∑pmin(n,m)∑k=1min(n,m)μ(k)F(pk)=∑pmin(n,m)∑k=1min(n,m)μ(k)⌊npk⌋⌊mpk⌋

直接这样做会TLE，那么考虑优化，令t=pk

ans=∑t=1min(m,n)∑p|t⌊nt⌋⌊mt⌋μ(tp)

我们便可以将a[t]=∑p|tμ(tp)预处理出来，再对其求前缀和，然后对其分块求解

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAXN 10000000#define LL long longbool check[MAXN+10];int primer[MAXN+10];int mu[MAXN+10];int sum[MAXN+10];void Moblus(){    memset(check,0,sizeof(check));    mu[1]=1;    int tot=0;    for(int i=2;i<=MAXN;i++){        if(!check[i]){            primer[tot++]=i;            mu[i]=-1;        }        for(int j=0;j<tot&&i*primer[j]<=MAXN;j++){            check[i*primer[j]]=true;            if(i%primer[j]==0){                mu[i*primer[j]]=0;                break;            }            mu[i*primer[j]]=-mu[i];        }    }    ///    sum[0]=0;    for(int i=0;i<tot;i++)        for(int j=primer[i];j<MAXN;j+=primer[i])            sum[j]+=mu[j/primer[i]];    for(int i=1;i<MAXN;i++)        sum[i]+=sum[i-1];}LL solve(int n,int m){    LL ans=0;    if(n>m) swap(n,m);    for(int i=1,last;i<=n;i=last+1){        last=min(n/(n/i),m/(m/i));        ans+=(LL)(sum[last]-sum[i-1])*(n/i)*(m/i);    }    return ans;}int main(){    int T;    Moblus();    int a,b;    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&a,&b);        LL ans=solve(a,b);        printf("%lld\n",ans);    }    return 0;}