BZOJ3994: [SDOI2015] Sum of approximations

BZOJ3994: [SDOI2015] Sum of approximations

Description

 Let d(x) be the number of divisors of x, given N and M, find $$ \sum_{i=1}^N\sum_{j=1}^Md(ij)$$

Input

The input file contains multiple sets of test data.

The first line, an integer T, represents the number of groups of test data.

The next T lines, each line with two integers N, M.

Output

 T lines, each with an integer representing the answer you are looking for.

Sample Input

2
7 4
5 6

Sample Output

110
121

HINT

1<=N, M<=50000

1<=T<=50000

 

Problem solving Here!

When I saw two summation symbols, I immediately thought of Mobius inversion.

But what about that d(ij)?

No problem, we have: d(ij)=\sum_{x|i}\sum_{y|j}[gcd(x,y)==1]

so,

ans=\sum_{i=1}^n\sum_{j=1}^m\sum_{u|i}\sum_{v|j}[gcd(u,v)==1]

　   =\sum_{i=1}^n\sum_{j=1}^m[\frac{n}{i}][\frac{m}{j}][gcd(i,j)==1]

 

Not finished. . .

Attached code:

#include<iostream>
#include<algorithm>
#include<cstdio>
#define MAXN 50010
using namespace std;
int k=0,prime[MAXN],mu[MAXN],sum[MAXN];
long long f[MAXN];
bool np[MAXN];
inline int read(){
	int date=0,w=1;char c=0;
	while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}
	while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}
	return date*w;
}
void make(){
	int m=MAXN-10;
	mu [1] = 1;
	for(int i=2;i<=m;i++){
		if(!np[i]){
			prime [++ k] = i;
			mu [i] = - 1;
		}
		for(int j=1;j<=k&&prime[j]*i<=m;j++){
			np[prime[j]*i]=true;
			if(i%prime[j]==0)break;
			else mu [prime [j] * i] = - mu [i];
		}
	}
	for(int i=1;i<=m;i++)sum[i]=sum[i-1]+mu[i];
	for(int i=1;i<=m;i++){
	    long long s=0;
	    for(int j=1,last=1;j<=i;j=last+1){
	        last=i/(i/j);
	        s+=(long long)(last-j+1)*(i/j);
	    }
	    f[i]=s;
	}
}
void work(){
    int n,m;
	long long ans=0;
	n=read();m=read();
	if(n>m)swap(n,m);
	for(int i=1,last=1;i<=n;i=last+1){
		last=min(n/(n/i),m/(m/i));
		ans+=(long long)(sum[last]-sum[i-1])*f[n/i]*f[m/i];
	}
	printf("%lld\n",ans);
}
int main(){
	int t=read();
	make();
	while(t--)work();
	return 0;
}