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

本文介绍了解决SPOJ平台上的PGCD问题的方法，该问题是关于计算一个由最大公约数构成的表格中素数出现次数的问题。通过构建一个高效的算法来遍历指定范围内的所有整数对，并统计当两个数的最大公约数为素数的情况。

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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，仅仅只是 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

具体解释：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;
}