Co-prime

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

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

Hint

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

分析：求出n的素因子，然后容斥求解出不互质的个数，剩下的就是互质的个数；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
int n,m,k,t,cnt,fac[maxn],cas;
ll x,y;
void init(int x)
{
    cnt=0;
    if(x%2==0){
        fac[++cnt]=2;
        while(x%2==0)x/=2;
    }
    for(int i=3;(ll)i*i<=x;i+=2)
    {
        if(x%i==0)
        {
            fac[++cnt]=i;
            while(x%i==0)x/=i;
        }
    }
    if(x>1)fac[++cnt]=x;
}
ll gao(ll x)
{
    ll ret=0;
    for(int i=1;i<(1<<cnt);i++)
    {
        ll num=0,now=1;
        for(int j=0;j<cnt;j++)
        {
            if(i&(1<<j))
            {
                ++num;
                now*=fac[j+1];
            }
        }
        if(num&1)ret+=x/now;
        else ret-=x/now;
    }
    return x-ret;
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%d",&x,&y,&n);
        init(n);
        printf("Case #%d: %lld\n",++cas,gao(y)-gao(x-1));
    }
    return 0;
}