SPOJ4491. Primes in GCD Table（gcd(a,b)=d素数，（1<=a<=n,1<=b<=m））加强版

SPOJ4491. Primes in GCD Table   Problem code: PGCD

 

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:

2
10 10
100 100

Output:

30
2791

 

 

一样的题，只不过 GCD(x,y) = 素数 .  1<=x<=a ; 1<=y<=b;

链接：http://www.spoj.com/problems/PGCD/ 

转载请注明出处：寻找&星空の孩子

 

详解：http://download.csdn.net/detail/u010579068/9034969

 

 

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn=1e7+5;
typedef long long LL;
LL pri[maxn],pnum;
LL mu[maxn];
LL g[maxn];
LL sum[maxn];
bool vis[maxn];

void mobius(int N)
{
    LL i,j;
    pnum=0;
    memset(vis,false,sizeof(vis));
    vis[1]=true;
    mu[1]=1;
    for(i=2; i<=N; i++)
    {
        if(!vis[i])//pri
        {
            pri[pnum++]=i;
            mu[i]=-1;
            g[i]=1;
        }
        for(j=0; j<pnum && i*pri[j]<=N ; j++)
        {
            vis[i*pri[j]]=true;
            if(i%pri[j])
            {
                mu[i*pri[j]]=-mu[i];
                g[i*pri[j]]=mu[i]-g[i];
            }
            else
            {
                mu[i*pri[j]]=0;
                g[i*pri[j]]=mu[i];
                break;//think...
            }
        }
    }
    sum[0]=0;
    for(i=1; i<=N; i++)
    {
        sum[i]=sum[i-1]+g[i];
    }
}
int main()
{
    mobius(10000000);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL n,m;
        scanf("%lld%lld",&n,&m);
        if(n>m) swap(n,m);
        LL t,last,ans=0;
        for(t=1;t<=n;t=last+1)
        {
            last = min(n/(n/t),m/(m/t));
            ans += (n/t)*(m/t)*(sum[last]-sum[t-1]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}