HDU-4135 Co-prime 容斥原理
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4135
题意：求区间[A,B]与K互素的数的个数。
首先对K分解质因数，然后容易原理搞，复杂度O(sqrt K)..
1 //STATUS:C++_AC_0MS_228KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=1000010; 37 const int INF=0x3f3f3f3f; 38 const int MOD=100000,STA=8000010; 39 const LL LNF=1LL<<60; 40 const double EPS=1e-8; 41 const double OO=1e15; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 LL A,B,K; 59 LL fac[50]; 60 61 LL solve(LL n,LL a){ 62 LL i,j,up,t,cnt=0,sum=0,flag; 63 for(i=2;i*i<=a;i++) 64 if(a%i==0){ 65 fac[cnt++]=i; 66 while(a%i==0)a/=i; 67 } 68 if(a>1)fac[cnt++]=a; 69 up=1<<cnt; 70 for(i=1;i<up;i++){ //容斥原理，二进制枚举 71 flag=0,t=1; 72 for(j=0;j<cnt;j++){ 73 if(i&(1<<j)){ 74 flag^=1; 75 t*=fac[j]; 76 } 77 } 78 sum+=flag?n/t:-(n/t); 79 } 80 return n-sum; 81 } 82 83 int main(){ 84 // freopen("in.txt","r",stdin); 85 int T,ca=1,i,j; 86 scanf("%d",&T); 87 while(T--) 88 { 89 scanf("%I64d%I64d%I64d",&A,&B,&K); 90 printf("Case #%d: %I64d\n",ca++,solve(B,K)-solve(A-1,K)); 91 } 92 93 return 0; 94 }
