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

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本文介绍了解决SPOJ4491.PrimesinGCDTable问题的方法，该问题是关于在一个由最大公约数构成的表格中计算素数出现次数的算法挑战。通过构建一个高效算法，文章详细解释了如何预先计算素数，并使用Mobius函数和前缀和技巧来快速得出答案。

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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.youkuaiyun.com/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;
}