hdu 4135 A-B中，与N不互质的数

传送门

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5403    Accepted Submission(s): 2151


Problem Description

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.

Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

 

Input

The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).

 

Output

For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

 

Sample Input

2
1 10 2
3 15 5

 

Sample Output

Case #1: 5
Case #2: 10

HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

 

Source

The Third Lebanese Collegiate Programming Contest

 

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求A-B中，与N不互质的数。也就是说求1-B中与N互质的数。再求1-（A-1）中与N互质的数，用B-（A-1）-互质的数=不互质的数。但有一个容斥在里面，因为N可以分解，

p1^a1*p2^a2*p3^a3````。奇加偶减。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e4+10;
const int maxx=1e6+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
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));}
inline LL Scan()
{
LL Res=0,ch,Flag=0;
if((ch=getchar())=='-')Flag=1;
else if(ch>='0' && ch<='9')Res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0';
return Flag ? -Res : Res;
}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.out" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int tot=0,cnt=0,vis[maxx];
LL pr[maxx],a[maxx/100];
LL A,B;
int N;
void init()
{
tot=0;
int len = sqrt(maxx) + 1;
for (int i = 2; i <= len; ++i)if (!vis[i])
{
pr[tot++]=i;
for (int j = i*i; j <= maxx; j += i)vis[j] = 1;
}
}
void pri(LL x)
{
cnt=0;
for(int i=0;i<tot&&pr[i]*pr[i]<=x;i++)
{
if(x%pr[i]==0)
{
a[cnt++]=pr[i];
while(x%pr[i]==0)
x/=pr[i];
}
}
if(x>1)
a[cnt++]=x;
}
LL solve(LL x)
{
LL st=(1<<cnt);
LL ret=0;
for(LL i=1;i<st;i++)
{
LL t=i,k=0,tmp=1;
for(LL j=0;j<cnt;j++)
{
//cout<<t<<endl;
if(t&1)
{

4000
k++;
tmp*=a[j];
}
t>>=1;
}
if(k&1)
ret+=x/tmp;
else ret-=x/tmp;
}
return ret;
}
int main()
{
init();
int t;
t=Scan();
for(int cas=1;cas<=t;cas++)
{
A=Scan();B=Scan();N=Scan();
pri(N);
LL ans=B-solve(B)-A+1+solve(A-1);
printf("Case #%d: %lld\n",cas,ans);
}
}