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

Posted llguanli

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了SPOJ4491. Primes in GCD Table(gcd(a,b)=d素数,(1<=a<=n,1<=b<=m))加强版相关的知识,希望对你有一定的参考价值。

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.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;
}


 

 

 

 

 

以上是关于SPOJ4491. Primes in GCD Table(gcd(a,b)=d素数,(1&lt;=a&lt;=n,1&lt;=b&lt;=m))加强版的主要内容,如果未能解决你的问题,请参考以下文章

SPOJ-PGCD Primes in GCD Table

SPOJ Distinct Primes 打表

bzoj2226[Spoj 5971] LCMSum

SPOJ HG - HUGE GCD

SPOJ:Bits. Exponents and Gcd(组合数+GCD)

为啥我的 SPOJ GCD2 代码在 SPOJ 上出错?