HDU 4135 Co-prime 容斥

http://acm.hdu.edu.cn/showproblem.php?pid=4135

因为要与n互素, 所以n的素因子显然是不能出现的...那么我们只要找出1-a和1-b中含n的素因子的数的个数即可...

显然会有大量重复,大胆容斥O(2^cnt)即可..cnt是n的素因子的个数

bitset维护所有素因子的二进制值,枚举一下就出来了...素因子不是很大...总体大概O(sqrt(n) * 2^(cnt)) 

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define EPS         1e-8
#define DINF        1e15
#define MAXN        105000
#define LINF        1LL << 60
#define MOD         1000000007
#define INF         0x7fffffff
#define PI          3.14159265358979323846
#define lson            l,m,rt<<1
#define rson            m+1,r,rt<<1|1
#define BUG             cout<<" BUG! "<<endl;
#define LINE            cout<<" ------------------ "<<endl;
#define FIN             freopen("in.txt","r",stdin);
#define FOUT            freopen("out.txt","w",stdout);
#define mem(a,b)        memset(a,b,sizeof(a))
#define FOR(i,a,b)      for(int i = a ; i < b ; i++)
#define read(a)         scanf("%d",&a)
#define read2(a,b)      scanf("%d%d",&a,&b)
#define read3(a,b,c)    scanf("%d%d%d",&a,&b,&c)
#define write(a)        printf("%d\n",a)
#define write2(a,b)     printf("%d %d\n",a,b)
#define write3(a,b,c)   printf("%d %d %d\n",a,b,c)
#pragma comment         (linker,"/STACK:102400000,102400000")
template<class T> inline T L(T a)       {return (a << 1);}
template<class T> inline T R(T a)       {return (a << 1 | 1);}
template<class T> inline T lowbit(T a)  {return (a & -a);}
template<class T> inline T Mid(T a,T b) {return ((a + b) >> 1);}
template<class T> inline T gcd(T a,T b) {return b ? gcd(b,a%b) : a;}
template<class T> inline T lcm(T a,T b) {return a / gcd(a,b) * 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));}
template<class T> inline T exGCD(T a, T b, T &x, T &y){
    if(!b) return x = 1,y = 0,a;
    T res = exGCD(b,a%b,x,y),tmp = x;
    x = y,y = tmp - (a / b) * y;
    return res;
}
//typedef long long LL;    typedef unsigned long long ULL;
typedef __int64 LL;      typedef unsigned __int64 ULL;
/*********************   By  F   *********************/

LL a,b,n,ans;
struct node{
    LL num;
    int cnt;
}prim[MAXN];
int cnt;

// O(sqrt(n)) 得到n的因子
void solve_n(LL n){
    cnt = 0;
    LL m = n;
    for(LL i = 2 ; i*i <= m; i++){
        if(n%i == 0){
            prim[cnt].num = i;
            while(n%i == 0){
                prim[cnt].cnt++;
                n/=i;
            }
            cnt++;
        }
    }
    if(n > 1) {
        prim[cnt].num = n;
        prim[cnt++].cnt = 1;
    }
    //for(int i = 0 ; i < cnt ; i++) cout<<prim[i].num<<" "<<prim[i].cnt<<endl;
}
// 容斥1-a中与n互素的数的个数
LL getprim(LL a,LL n){
    LL ret = 0;
    for(int i = 1 ; i < (1<<cnt) ; i++){
        bitset<32> bit(i);
        int p = bit.count(),mul = 1;
        for(int i = 0 ; i < 32 ; i++)
            if(bit[i]) mul *= prim[i].num;
        if(p&1) ret += (a/mul);
        else ret -= (a/mul);
    }
    return a - ret;
}
int main(){
    //FIN;
    //FOUT;
    int T;
    scanf("%d",&T);
    for(int cas = 1 ; cas <= T ;cas++){
        mem(prim,0);
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        solve_n(n);
        LL ans = getprim(b,n) - getprim(a,n);
        if(gcd(a,n)==1) ans++;
        printf("Case #%d: %I64d\n",cas,ans);
    }
    return 0;
}