容斥原理+质因数分解-HDU4135

https://vjudge.net/problem/HDU-4135

掌握质因数分解和容斥原理的应用

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int inf=10056;
const int maxn=1;
ll a,b;
int n;
int num;
ll fen[100];
ll gcd(ll a,ll b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);
}
ll ge(ll x)
{
    ll sum=0;
    for(int i=1;i<(1<1;int k=0;
        for(int j=0;jif(i&(1<if(k&1)
            sum=sum+x/lcm;
        else
            sum=sum-x/lcm;
    }
    return sum;
}
int main()
{
    int t;
    scanf("%d",&t);
    int qqq=1;
    while(t--)
    {
        scanf("%I64d %I64d %d",&a,&b,&n);
        int tag=1;
        int s=sqrt(n);
        num=0;
        for(int i=2;i<=n;i++)//质因数分解
        {
            if(n%i==0)
            {
                fen[num++]=i;
                while(n%i==0)
                    n=n/i;
                tag=0;
            }
            if(i>s&&tag)
            {
                fen[num++]=n;
                break;
            }

        }
        ll s1=ge(b);
        ll s2=ge(a-1);
        //cout<
        ll ans=(b-a+1)-(s1-s2);
        printf("Case #%d: %I64d\n",qqq++,ans);


    }
    return 0;
}