HDU-4135 Co-prime 容斥原理
2013-08-19 23:24
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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)..
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)..
//STATUS:C++_AC_0MS_228KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cstring> #include <cassert> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <list> #include <set> #include <map> using namespace std; //#pragma comment(linker,"/STACK:102400000,102400000") //using namespace __gnu_cxx; //define #define pii pair<int,int> #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define PI acos(-1.0) //typedef typedef __int64 LL; typedef unsigned __int64 ULL; //const const int N=1000010; const int INF=0x3f3f3f3f; const int MOD=100000,STA=8000010; const LL LNF=1LL<<60; const double EPS=1e-8; const double OO=1e15; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //Daily Use ... inline int sign(double x){return (x>EPS)-(x<-EPS);} template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} template<class T> inline T lcm(T a,T b,T d){return a/d*b;} template<class T> inline T Min(T a,T b){return a<b?a:b;} template<class T> inline T Max(T a,T b){return a>b?a:b;} template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} //End LL A,B,K; LL fac[50]; LL solve(LL n,LL a){ LL i,j,up,t,cnt=0,sum=0,flag; for(i=2;i*i<=a;i++) if(a%i==0){ fac[cnt++]=i; while(a%i==0)a/=i; } if(a>1)fac[cnt++]=a; up=1<<cnt; for(i=1;i<up;i++){ //容斥原理,二进制枚举 flag=0,t=1; for(j=0;j<cnt;j++){ if(i&(1<<j)){ flag^=1; t*=fac[j]; } } sum+=flag?n/t:-(n/t); } return n-sum; } int main(){ // freopen("in.txt","r",stdin); int T,ca=1,i,j; scanf("%d",&T); while(T--) { scanf("%I64d%I64d%I64d",&A,&B,&K); printf("Case #%d: %I64d\n",ca++,solve(B,K)-solve(A-1,K)); } return 0; }
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