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

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 #define MAX 100010
 6 #define ll long long
 7 ll mu[MAX]= {0},mb[MAX]= {0};
 8 void init()
 9 {
10     int i,j;
11     for(i=2; i<MAX; i++)
12     {
13         if(!mu[i])
14         {
15             mu[i]=i;
16             if(i>1000)continue;
17             j=i*i;
18             while(j<MAX)
19             {
20                 mu[j]=i;
21                 j+=i;
22             }
23         }
24     }
25     mu[1]=1;
26     for(i=2; i<MAX; i++)
27     {
28         if((i/mu[i])%mu[i]==0)mu[i]=0;
29         else mu[i]=-mu[i/mu[i]];
30     }
31     for(i=1; i<MAX; i++)
32     mb[i]=mu[i]*i,mu[i]+=mu[i-1],mb[i]+=mb[i-1];
33 }
34 void solve(int n,int m)
35 {
36     ll ans,ansx,ansy,x,y;
37     ans=ansx=ansy=0;
38     int i,j,k;
39     for(i=(n>m?m:n);i>0;)
40     {
41         x=n/i,y=m/i;
42         k=max(n/(x+1),m/(y+1));
43         ans+=x*y*(mu[i]-mu[k]);
44         ansx+=(mb[i]-mb[k])*y*(1+x)*x/2;
45         ansy+=(mb[i]-mb[k])*x*(1+y)*y/2;
46         i=k;
47     }
48     printf("%lld %lld %lld\n",ans,ansx,ansy);
49 }
50 int main()
51 {
52     init();
53     int n,m;
54     while(scanf("%d%d",&n,&m)!=EOF)
55     {
56         solve(n,m);
57     }
58 }

GCD SUM 强大的数论，容斥定理