数学: HDU Co-prime

Co-prime

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 4

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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

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}.

Source

The Third Lebanese Collegiate Programming Contest

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#include<iostream> #include<cstring> #include<cstdio>    using namespace std; typedef long long LL; const int N = 1e5+5;    LL f[N],prime[N],vis[N],cnt,k; void prime_factor(){     memset(vis,0,sizeof(vis));     vis[0]=vis[1] = 1,cnt = 0;     for(LL i=2;i*i<N;i++)     if(!vis[i]) for(LL j=i*i;j<N;j+=i) vis[j] = 1;     for(LL i=0;i<N;i++) if(!vis[i]) prime[cnt++] = i; } LL poie(LL x){     LL ret = 0,sum,tmp;     for(LL i=1;i<(1LL<<k);i++){         tmp = 1,sum=0;         for(LL j=0;j<k;j++) if(i&(1LL<<j)){sum++,tmp*=f[j];}         if(sum&1) ret += x/tmp;         else ret -= x/tmp;     }     return ret; }    void solve_question(LL A,LL B,LL n){     LL tmp = n;     k = 0 ;     for(LL i=0;prime[i]*prime[i]<= tmp;i++){         if(tmp%prime[i]==0)             f[k++] = prime[i];         while(tmp%prime[i]==0)             tmp/=prime[i];     }     if(tmp > 1) f[k++] = tmp;     LL ans =B-poie(B)-A+1+poie(A-1);     printf("%I64d\n",ans); } int main(){     int T,Case=0;     LL A,B,n;     scanf("%d",&T);     prime_factor();     while(T--){         scanf("%I64d %I64d %I64d",&A,&B,&n);         printf("Case #%d: ",++Case);         solve_question(A,B,n);     } }