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]= {},mb[MAX]= {};
 void init()
 {
     int i,j;
     for(i=; i<MAX; i++)
     {
         if(!mu[i])
         {
             mu[i]=i;
             if(i>)continue;
             j=i*i;
             while(j<MAX)
             {
                 mu[j]=i;
                 j+=i;
             }
         }
     }
     mu[]=;
     for(i=; i<MAX; i++)
     {
         if((i/mu[i])%mu[i]==)mu[i]=;
         else mu[i]=-mu[i/mu[i]];
     }
     for(i=; i<MAX; i++)
     mb[i]=mu[i]*i,mu[i]+=mu[i-],mb[i]+=mb[i-];
 }
 void solve(int n,int m)
 {
     ll ans,ansx,ansy,x,y;
     ans=ansx=ansy=;
     int i,j,k;
     for(i=(n>m?m:n);i>;)
     {
         x=n/i,y=m/i;
         k=max(n/(x+),m/(y+));
         ans+=x*y*(mu[i]-mu[k]);
         ansx+=(mb[i]-mb[k])*y*(+x)*x/;
         ansy+=(mb[i]-mb[k])*x*(+y)*y/;
         i=k;
     }
     printf("%lld %lld %lld\n",ans,ansx,ansy);
 }
 int main()
 {
     init();
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         solve(n,m);
     }
 }