Hillan and the girl

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Problem Description

“WTF!
While everyone has his girl(gay) friend, I only have my keyboard!”
Tired of watching others‘ affair, Hillan burst into scream, which made
him decide not to hold it back.
“All right, I am giving you a
question. If you answer correctly, I will be your girl friend.” After
listening to Hillan, Girl replied, “What is the value of ∑ni=1∑mj=1f(i,j), where f(i,j)=0 if gcd(i,j) is a square number and f(i,j)=1 if gcd(i,j) is not a square number(gcd(i,j) means the greatest common divisor of x and y)?”
But Hillan didn‘t have enough Intelligence Quotient to give the right answer. So he turn to you for help.

Input

The first line contains an integer T(1≤T≤10,000)——The number of the test cases.
For each test case, the only line contains two integers n,m(1≤n,m≤10,000,000) with a white space separated.

Output

For each test case, the only line contains a integer that is the answer.

Sample Input

2
1 2333333
10 10

Sample Output

0
33

Hint

In the first test case, obviously $f\left(i,j\right)$ always equals to 0, because $i$ always equals to 1 and $\gcd\left(i,j\right)$ is always a square number(always equals to 1).

Source

BestCoder Round #79 (div.2)

min(n,m) min(n/k,m/k)

思路:首先推到∑ 　　　　∑　mu(d)  * [n/k/d] * [m/k/d];  k为完全平方数；

　　　　　　k=1　　　d=1

　　　令T=k*d;

　　　可得:

　　　　　　min(n,m)　　　　　　　　　　　　　　　　　　　　

　　　　　　∑　　　[n/T] * [m/T]  ∑   mu(T/k)  ;

　　　　　　T　　　　　　　　　　k|T

　　　　　　令gg数组等于 ∑   mu(T/k)  相当于原来的mu函数;

　　　　　　　　　　　　 k|T

　　　　　　和原来一样分块即可；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
#define pi 4*atan(1)
const int N=1e7+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
int mu[N], p[N], np[N], cnt, sum[N];
ll gg[N];
void init() {
    mu[1]=1;
    for(int i=2; i<N; ++i) {
        if(!np[i]) p[++cnt]=i, mu[i]=-1;
        for(int j=1; j<=cnt && i*p[j]<N; ++j) {
            int t=i*p[j];
            np[t]=1;
            if(i%p[j]==0) { mu[t]=0; break; }
            mu[t]=-mu[i];
        }
    }
    for(int i=1;i*i<N;i++)
    {
        for(int t=i*i;t<N;t+=(i*i))
        sum[t]+=mu[t/i/i];
    }
    for(int i=1;i<N;i++)
    gg[i]=gg[i-1]+sum[i];

}
ll getans(ll b,ll d)
{
    if(b>d)swap(b,d);
    ll ans=0;
    for(ll L=1,R=0;L<=b;L=R+1)
    {
        R=min(b/(b/L),d/(d/L));
        ans+=(b/L)*(d/L)*(gg[R]-gg[L-1]);
    }
    return ans;
}
int main()
{
    int T;
    init();
    scanf("%d",&T);
    while(T--)
    {
        ll b,d;
        scanf("%I64d%I64d",&b,&d);
        printf("%I64d\n",(b*d)-getans(b,d));
    }
    return 0;
}