【BZOJ2820】YY的GCD [莫比乌斯反演]
YY的GCD
Time Limit: 10 Sec Memory Limit: 512 MB[Submit][Status][Discuss]
Description
求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对k。
Input
第一行一个整数T 表述数据组数接下来T行，每行两个正整数，表示N, M。
Output
T行，每行一个整数表示第 i 组数据的结果
Sample Input
2
10 10
100 100
10 10
100 100
Sample Output
30
2791
2791
HINT
T = 10000
N, M <= 10000000
Solution
Code
1 #include<iostream>
2 #include<string>
3 #include<algorithm>
4 #include<cstdio>
5 #include<cstring>
6 #include<cstdlib>
7 #include<cmath>
8 using namespace std;
9 typedef long long s64;
10
11 const int ONE = 10000005;
12
13 int T;
14 int n,m;
15 bool isp[ONE];
16 int prime[700005],p_num;
17 int miu[ONE],sum[ONE];
18 s64 Ans;
19
20 int get()
21 {
22 int res=1,Q=1; char c;
23 while( (c=getchar())<48 || c>57)
24 if(c=='-')Q=-1;
25 if(Q) res=c-48;
26 while((c=getchar())>=48 && c<=57)
27 res=res*10+c-48;
28 return res*Q;
29 }
30
31 void Getmiu(int MaxN)
32 {
33 miu[1] = 1;
34 for(int i=2; i<=MaxN; i++)
35 {
36 if(!isp[i])
37 prime[++p_num] = i, miu[i] = -1;
38 for(int j=1; j<=p_num, i*prime[j]<=MaxN; j++)
39 {
40 isp[i * prime[j]] = 1;
41 if(i % prime[j] == 0)
42 {
43 miu[i * prime[j]] = 0;
44 break;
45 }
46 miu[i * prime[j]] = -miu[i];
47 }
48 }
49 for(int j=1; j<=p_num; j++)
50 for(int i=1; i*prime[j]<=MaxN; i++)
51 sum[i * prime[j]] += miu[i];
52 for(int i=1; i<=MaxN;i++)
53 sum[i] += sum[i-1];
54 }
55
56 void Solve()
57 {
58 n=get(); m=get();
59 if(n > m) swap(n,m);
60 Ans = 0;
61 for(int i=1, j=0; i<=n; i=j+1)
62 {
63 j = min(n/(n/i), m/(m/i));
64 Ans += (s64) (n/i) * (m/i) * (sum[j] - sum[i-1]);
65 }
66 printf("%lld\n",Ans);
67 }
68
69 int main()
70 {
71 Getmiu(ONE-1);
72 T=get();
73 while(T--)
74 Solve();
75 }
