Hdu 4497 GCD and LCM decomposition of number theory primes

GCD and LCMTime limit:2000/1000 MS (java/others) Memory limit:65535/65535 K (java/others)
Total submission (s): 1339 Accepted Submission (s): 607


Problem Descriptiongiven-positive integers G and L, could you tell me how many solutions of (x, Y, z) there is, Satis Fying that gcd (x, y, z) = G and LCM (x, y, z) = L?
Note, gcd (x, y, z) means the greatest common divisor of x, Y and Z, while LCM (x, y, z) means the least common multiple of x, y, and Z.
Note 2, (1, 2, 3) and (1, 3, 2) are different solutions.
Inputfirst line comes an integer t (t <=), telling the number of test cases.
The next T lines, each contains, positive 32-bit signed integers, G and L.
It's guaranteed that each answer would fit in a 32-bit signed integer.
Outputfor each test case, print one line with the number of solutions satisfying the conditions above.
Sample Input

Sample Output

72 0

Links: http://acm.hdu.edu.cn/showproblem.php?pid=4497

Test instructions: Each case gives you two numbers G and L. Then find out how many different (x, Y, z) gcd are G, and the LCM is L.

Procedure: Decomposition of prime g,l.

First l%g! =0 that must be output 0.

First case 6 72

After decomposition 6 has a 2, a 3.

72 after decomposition there are 3 2, 2 3.

Then for the prime number 2, 6 of a 2 is the lower limit, and 72 of 3 2 is the upper limit. X, Y, z must have a number to decompose a prime 2, must have another number to decompose 3 2, and then the remaining number of the 2 must be divided between "1,3".

Then it can be introduced for the number of prime 2 to X, Y, Z assigned primes 2 Total has 6+ (3-1-1) *6 species Division. The preceding 6, which represents the third number, is equal to the upper or lower bounds such as 2 2 8 or 2 8 8.  Since there are two numbers equal, their permutation combinations are 3 kinds.   So it's 2*3=6 species.   Then the following is the case of three numbers are different, 3-1-1 =1 1 to 3 is a number, that is, 2, that is, the allocation of two prime number 2. Then 2 4 8 are arranged in combination, three are different, and the permutations are grouped into 6.

So there are 12 kinds of allocation methods for prime number 2.

Then allocate 3, a total of 6.

Multiply the number of allocation methods for each prime 12*6=72. Is the total number of programs.

Obviously for this problem, can only use decomposition l/g, and then for L each prime number of the lower limit is all 0.

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> #include <malloc.h> #include <ctype.h> #include <math.h># Include <string> #include <iostream> #include <algorithm>using namespace std; #include <stack># Include <queue> #include <vector> #include <deque> #include <set> #include <map> #define LL __int64__int64 gcd (__int64 A,__int64 b) {if (b==0) return A;return gcd (b,a%b);} __int64 LCM (__int64 A,__int64 b) {return a/gcd (b) *b;} #define N 10000000//prime[0,primenum) ll prime[1000000], Primenu  M  BOOL ISPRIME[N+10];     <=max_prime prime void Prime (ll max_prime) {primenum = 0;      Isprime[0] = isprime[1] = 0;      ISPRIME[2] = 1;      prime[primenum++] = 2;      for (ll i = 3; I <= max_prime; i++) isprime[i] = i&1; for (ll i = 3; I <= max_prime; i+=2) {if (Isprime[i]) prime[primenum++] = i;              for (ll j = 0; J < Primenum; J + +) {if (prime[j] * i > Max_prime) break;              Isprime[prime[j]*i] = 0;          if (i%prime[j] = = 0) break;  }}} ll fen_g[1000000];ll fen_l[1000000];ll vis[1000000];ll wei[1000000];int Main () {PRIME (100000); __int64 t,g,l; scanf ("%i64d", &t), while (t--) {scanf ("%i64d%i64d", &g,&l), if (l%g!=0) printf ("0\n"), else {memset (fen_g,0 , sizeof fen_g); memset (fen_l,0,sizeof fen_l); memset (vis,0,sizeof vis); L=l/g;__int64 Tem_g=g;__int64 Tem_l=l;__int64 Ji=0;for (__int64 i=0;i<primenum;i++) {while (tem_l%prime[i]==0) {tem_l/=prime[i];if (Fen_l[i]==0&&vis[i] ==0) {vis[i]=1;wei[ji++]=i;}  fen_l[i]++;//Upper Limit}}if (tem_l!=1) {wei[ji++]=primenum;vis[primenum]=1;fen_l[primenum]++;} ll Ans=1;for (__int64 i=0;i<ji;i++) {__int64 ww=wei[i]; __int64 Cnt=fen_l[wei[i]]; if (cnt==1) {ans*=6;} elseans*=3*2+ (cnt-1) *6; }printf ("%i64d\n", ans);}} return 0;  }

Hdu 4497 GCD and LCM decomposition of number theory primes