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)..

//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
",ca++,solve(B,K)-solve(A-1,K));
    }

    return 0;
}