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.
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
" ,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);
}
}
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