题意：http://acm.hdu.edu.cn/showproblem.php?pid=2841
找到x，y互质的坐标个数
思路：
和hdu4135很类似。
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> HDU2841 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 38 #define ls rt<<1 39 #define rs rt<<1|1 40 typedef pair<int,int> PII; 41 typedef vector<int> VI; 42 typedef unsigned long long ull; 43 typedef long long ll; 44 typedef double db; 45 const db E=2.718281828; 46 const db PI=acos(-1.0); 47 const ll INF=(1LL<<60); 48 const int inf=(1<<30); 49 const db ESP=1e-9; 50 const int mod=(int)1e9+7; 51 #define int ll 52 const int N=(int)1e6+10; 53 54 int pri[50],num; 55 void getpri(int n) 56 { 57 num=0; 58 for(int i=2;i*i<=n;++i) 59 { 60 if(n%i)continue; 61 while(n%i==0)n/=i; 62 pri[++num]=i; 63 } 64 if(n>1) 65 pri[++num]=n; 66 } 67 68 int getnocoprime(int x) 69 { 70 int ans=0; 71 int tot=(1<<num)-1; 72 for(int i=1;i<=tot;++i) 73 { 74 int sum=0; 75 int temp=1; 76 for(int j=0;j<=num-1;++j) 77 { 78 if((i>>j)&1) 79 sum++,temp*=pri[j+1]; 80 } 81 if(sum&1) 82 ans+=x/temp; 83 else 84 ans-=x/temp; 85 } 86 return ans; 87 } 88 89 void solve() 90 { 91 int n,m; 92 sc("%lld%lld",&n,&m); 93 int ans=0; 94 for(int i=1;i<=n;++i) 95 { 96 getpri(i); 97 ans+=m-getnocoprime(m); 98 } 99 pr("%lld ",ans); 100 } 101 102 signed main() 103 { 104 int T; 105 sc("%lld",&T); 106 while(T--)solve(); 107 return 0; 108 } 109 110 /**************************************************************************************/