hdu4135 Co-prime

Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6933    Accepted Submission(s): 2733   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 扫描二维码关注公众号，回复： 2503660 查看本文章   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}. 代码： #include<cstdio> #include<algorithm> #include<map> #define ll long long using namespace std; map<ll,ll> mp; ll a[2000];     int pos;  void solve(ll n) {     pos = 0;     for (ll i = 2;i * i <= n;i ++)     {         if (n % i == 0)         {             while (n % i == 0)             {                 a[pos] = i;                 n /= i;             }             pos ++;         }     }     if (n != 1) a[pos ++] = n; } ll exc(ll n,ll m) {     ll ans = 0;     for (int i = 1;i < (1 << m);i ++)     {         int cnt = 0,lm = 1;         for (int j = 0;j < m;j ++)         {             if ((i >> j) & 1)             {                 cnt += 1;                 lm *= a[j];             }         }         if (cnt & 1) ans += n / lm;         else ans -= n / lm;     }     return ans; } int main() {     int t,kase = 0;     scanf("%d",&t);     while (t--)     {         kase ++;         ll n,a,b;         scanf("%lld %lld %lld",&a,&b,&n);         solve(n);         ll numb = exc(b,pos);         ll numa = exc(a - 1,pos);         printf("Case #%d: %lld\n",kase,b - a + 1 - (numb - numa));     }     return 0; }