GCD SUM 强大的数论，容斥定理

GCD SUM

Time Limit: 8000/4000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

SubmitStatus

Problem Description

给出N,M
执行如下程序：
long long  ans = 0,ansx = 0,ansy = 0;
for(int i = 1; i <= N; i ++)
   for(int j = 1; j <= M; j ++)
       if(gcd(i,j) == 1) ans ++,ansx += i,ansy += j;
cout << ans << " " << ansx << " " << ansy << endl;

Input

多组数据，每行两个数N,M(1 <= N,M <= 100000)。

 

Output

如题所描述，每行输出3个数，ans,ansx,ansy，空格隔开

Sample Input

5 5
1 3

Sample Output

19 55 55
3 3 6

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAX 100010
#define ll long long
ll mu[MAX]= {0},mb[MAX]= {0};
void init()
{
    int i,j;
    for(i=2; i<MAX; i++)
    {
        if(!mu[i])
        {
            mu[i]=i;
            if(i>1000)continue;
            j=i*i;
            while(j<MAX)
            {
                mu[j]=i;
                j+=i;
            }
        }
    }
    mu[1]=1;
    for(i=2; i<MAX; i++)
    {
        if((i/mu[i])%mu[i]==0)mu[i]=0;
        else mu[i]=-mu[i/mu[i]];
    }
    for(i=1; i<MAX; i++)
    mb[i]=mu[i]*i,mu[i]+=mu[i-1],mb[i]+=mb[i-1];
}
void solve(int n,int m)
{
    ll ans,ansx,ansy,x,y;
    ans=ansx=ansy=0;
    int i,j,k;
    for(i=(n>m?m:n);i>0;)
    {
        x=n/i,y=m/i;
        k=max(n/(x+1),m/(y+1));
        ans+=x*y*(mu[i]-mu[k]);
        ansx+=(mb[i]-mb[k])*y*(1+x)*x/2;
        ansy+=(mb[i]-mb[k])*x*(1+y)*y/2;
        i=k;
    }
    printf("%lld %lld %lld
",ans,ansx,ansy);
}
int main()
{
    init();
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        solve(n,m);
    }
}