acdream 1148 GCD SUM 莫比乌斯反演 ansx,ansy

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 typedef long long LL;
 7 
 8 const int maxn = 1e5+3;
 9 bool s[maxn];
10 int prime[maxn],len = 0;
11 int mu[maxn];
12 LL hxl [maxn];
13 int sum1[maxn];
14 void  init()
15 {
16     memset(s,true,sizeof(s));
17     mu[1] = 1;
18     for(int i=2;i<maxn;i++)
19     {
20         if(s[i] == true)
21         {
22             prime[++len]  = i;
23             mu[i] = -1;
24         }
25         for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++)
26         {
27             s[i*prime[j]] = false;
28             if(i%prime[j]!=0)
29                 mu[i*prime[j]] = -mu[i];
30             else
31             {
32                 mu[i*prime[j]] = 0;
33                 break;
34             }
35         }
36     }
37     for(int i=1;i<maxn;i++)
38         sum1[i] = sum1[i-1]+mu[i];
39     hxl[1] = mu[1];
40     for(int i=2;i<maxn;i++){
41         hxl[i] = i*mu[i]+hxl[i-1];
42     }
43 }
44 int main()
45 {
46     init();
47     int n,m;
48     while(scanf("%d%d",&n,&m)>0)
49     {
50         LL sum = 0;
51         LL ansi = 0,ansj = 0;
52         int a = n;
53         int b = m;
54         if(a>b) swap(a,b);
55         for(int i=1,la = 0;i<=a;i++,i = la+1)
56         {
57             la = min(a/(a/i),b/(b/i));
58             sum = sum + ((LL)(a/i))*(b/i)*(sum1[la]-sum1[i-1]);
59             ansi = ansi +(hxl[la]-hxl[i-1])*(((LL)(n/i+1)*(n/i))/2)*(m/i);
60             ansj = ansj +(hxl[la]-hxl[i-1])*(((LL)(m/i+1)*(m/i))/2)*(n/i);
61         }
62         printf("%lld %lld %lld\n",sum,ansi,ansj);
63     }
64     return 0;
65 }