hdu1695

求两段区间中最大公约数为k的数字的对数，下界除以k后就会变成，求区间中与n互质的数字的对数。主要错在了两个地方，k=0和b，d大小的判断。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <sstream>
#include <fstream>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>

using namespace std;
typedef pair<int,int> II;
typedef vector<int> IV;
typedef vector<II> IIV;
typedef vector<bool> BV;
typedef long long i64;
typedef unsigned long long u64;
typedef unsigned int u32;
#define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)
#define IsComp(n) (_c[n>>6]&(1<<((n>>1)&31)))
#define SetComp(n) _c[n>>6]|=(1<<((n>>1)&31))
const int MAXP = 46341;
const int SQRP = 216;
int _c[(MAXP>>6)+1];
IV primes;
void prime_sieve() {
    for ( int i = 3; i <= SQRP; i += 2 )
        if ( !IsComp(i) ) for ( int j = i*i; j <= MAXP; j+=i+i ) SetComp(j);
    primes.push_back ( 2 );
    for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp(i) ) primes.push_back ( i );
}
void prime_factorize ( int n, IIV &f ) {
    int sn =(int) sqrt ( (double)n );
    For ( IV, it, primes ) {
        int prime = *it;
        if ( prime > sn ) break; if ( n % prime ) continue;
        int e = 0; for ( ; n%prime == 0; e++, n/= prime );
        f.push_back ( II(prime,e) );
        int sn = (int)sqrt( (double)n);
    }
    if ( n > 1 ) f.push_back( II(n,1) );
}
IIV aa;
u64 dfs ( int cur, int val ) {
    u64 res = 0;
    for ( int i = cur; i < aa.size(); ++i ) {
        res += val/aa[i].first- dfs( i+1,val/aa[i].first );
    }
    return res;
}
u64 fun ( int val, int m ) {
    aa.clear ( );
    prime_factorize ( m, aa );
    return val - dfs(0,val);
}
int main()
{
    prime_sieve ( );
    int a,b,c,d,n,sum,i,j,count=1,tmp,t;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>c>>d>>n;
        if(n==0)
        {
            cout<<"Case "<<count++<<": 0"<<"\n";
            continue;
        }
        if(b>d)
        {
            tmp=b;b=d;d=tmp;
        }
        b/=n;d/=n;
        sum=0;
        for(i=c;i<=d;i++)
        sum+=fun(min(i,b),i);
        cout<<"Case "<<count++<<": "<<sum<<"\n";
    }
    return 0;
}