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\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); }} |
