hdu 4135 Co-prime(素数分解 容斥)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<vector>
 5 
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 ll solve(ll a,ll n){
10     vector<ll> v;
11     for(ll i = 2 ; i*i <= n ; i ++){    // 质数分解 
12         if(n % i == 0){
13             v.push_back(i);
14             while(n%i == 0)    n/=i;
15         }
16     }
17     if(n > 1) v.push_back(n);
18     //有几位素因子 ,用二进制表示是否用过 
19     ll sum = 0,tmp,cnt;        
20     for(ll i = 1 ; i < (1<<v.size()) ; i ++){
21         cnt = 0;
22         tmp = 1;
23         for(ll j = 0 ; j < v.size() ; j ++){
24             if(i & (1<<j)){
25                 tmp *= v[j];    // 2 3;  6 12 18
26                 cnt++;
27             }
28         }
29         if(cnt & 1)  sum += a/tmp; //容斥 ,sum表示有1-a中有几个数字与n不互质 
30         else sum -= a/tmp;
31     }     
32     return a - sum; 
33 }
34 
35 int main(){
36     int T;
37     scanf("%d",&T);
38     for(int cas = 1 ; cas <= T ; cas ++){
39         ll a,b,n;
40         scanf("%lld%lld%lld",&a,&b,&n);
41         printf("Case #%d: %lld\n",cas,solve(b,n) - solve(a-1,n)); 
42     }
43     return 0;
44 }