hdu 4135 Co-prime 容斥原理

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}.

 

Source

The Third Lebanese Collegiate Programming Contest

题意：给你三个数，求[A,B]区间内与N互质的数的个数；

思路：即求[1,B]-[1,A-1]内与N互质的数

　　　因为n为1e9,所以根号打表，将N唯一分解，

　　　容斥得到与N不互质的数的个数；

　　　ans=B-[1,B]-(A-1-[1,A-1]);

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll prime[100010];
ll vis[100010];
ll a[110];
ll ji,cnt;
ll ans,x,y,z;
ll gcd(ll x,ll y)
{
    return y==0?x:gcd(y,x%y);
}
void Prime(ll n)
{
    cnt=0;
    memset(vis,0,sizeof(vis));
    for(ll i=2;i<n;i++)
    {
        if(!vis[i])
        prime[cnt++]=i;
        for(ll j=0;j<cnt&&i*prime[j]<n;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)//关键
            break;
        }
    }
}
void dfs(ll lcm,ll pos,ll step,ll x,ll &ans)
{
    if(lcm>x)
    return;
    if(pos==ji)
    {
        if(step==0)
            return;
        if(step&1)
        ans+=(x/lcm);
        else
        ans-=(x/lcm);
        return;
    }
    dfs(lcm,pos+1,step,x,ans);
    dfs(lcm/gcd(a[pos],lcm)*a[pos],pos+1,step+1,x,ans);
}
int main()
{
    ll t,i;
    int cs=1;
    Prime(100000);
    scanf("%I64d",&t);
    while(t--)
    {
        ji=0;
        scanf("%I64d%I64d%I64d",&x,&y,&z);
        for(i=0;i<cnt;i++)
        if(z%prime[i]==0)
        {
            a[ji++]=prime[i];
            while(z%prime[i]==0)
            z/=prime[i];
        }
        if(z>1)
        a[ji++]=z;
        ll gg=0;
        ans=0;
        dfs(1,0,0,x-1,gg);
        dfs(1,0,0,y,ans);
        printf("Case #%d: ",cs++);
        printf("%I64d
",(y-ans)-(x-1-gg));
    }
    return 0;
}