HDU 4135 Co-prime Euler+反発定理
10679 ワード
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 935 Accepted Submission(s): 339
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 <= 10
15) and (1 <=N <= 10
9).
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
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 935 Accepted Submission(s): 339
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 <= 10
15) and (1 <=N <= 10
9).
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
1 /*
2 ： [A,B], N 。
3
4 ：1. [A,B]，
5 [1，B] N
6
7 [1,A-1] N 。
8 [1，K] 。
9 2. [1，K] N K [1，K] N
10 gcd(N,k1)>1 。
11 ， N 。 ？？
12 。
13 。
14 ， ， 。
15 ， 。
16 3. 。
17 m=12,n=30
18 ： n ：2,3,5；
19 ：(1,m) n
20 (2,4,6,8,10)->n/2 6 ,(3,6,9,12)->n/3 4 ,(5,10)->n/5 2 。
21 ：6+4+2=12 ，
22 , ,
23 ！
24 ： ， ：n/2+n/3+n/5-n/(2*3)-n/(2*5)-n/(3*5)+n/(2*3*5).
25 : ？
26 ：dfs( )， ， ！
27 ：n ，n 。
28 ： -1 ！
29 */
30
31 #include<stdio.h>
32 #include<string.h>
33 #include<stdlib.h>
34
35 __int64 Que[100001];
36 __int64 f[100],len;
37 void Euler(__int64 n)// ， f[]
38 {
39 __int64 i;
40 len=0;
41 for(i=2;i*i<=n;i++)
42 {
43 if(n%i==0)
44 {
45 while(n%i==0)
46 n=n/i;
47 f[++len]=i;
48 }
49 }
50 if(n!=1)
51 f[++len]=n;
52 }
53
54 __int64 Capticy(__int64 num)
55 {
56 __int64 i,j,t=0,k,sum=0;
57 Que[t++]=-1;
58 for(i=1;i<=len;i++)
59 {
60 k=t;
61 for(j=0;j<k;j++)
62 Que[t++]=-1*Que[j]*f[i];
63 }
64 for(i=1;i<t;i++)
65 sum=sum+num/Que[i];
66 return sum;
67 }
68
69 int main()
70 {
71 __int64 T,A,B,N,i,k;
72 while(scanf("%I64d",&T)>0)
73 {
74 for(i=1;i<=T;i++)
75 {
76 scanf("%I64d%I64d%I64d",&A,&B,&N);
77 Euler(N);
78 k=B-Capticy(B)-(A-1-Capticy(A-1));
79 printf("Case #%I64d: %I64d
",i,k);
80 }
81 }
82 return 0;
83 }