2 1 10 2 3 15 5
Case #1: 5 Case #2: 10HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int mod = 1000000007;
LL a,b,n;
int t;
LL solve(LL r,LL n)
{
vector<int>v;
for(int i=2;i*i<=n;i++)
{
if(n&&n%i==0)
{
v.push_back(i);
while(n&&n%i==0)
n/=i;
}
}
if(n>1) v.push_back(n);
LL sum=0;
for(int t=1;t<(1<<v.size());t++)
{
LL mul=1,bits=0;
for(int i=0;i<(int)v.size();i++)
{
if(t&(1<<i))
{
++bits;
mul*=v[i];
}
}
if(bits&1) sum+=r/mul;
else sum-=r/mul;
}
return r-sum;
}
int main()
{
int cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d%I64d",&a,&b,&n);
printf("Case #%d: %I64d\n",cas++,solve(b,n)-solve(a-1,n));
}
return 0;
}
原文：http://blog.csdn.net/u013582254/article/details/42472307