HDU 4497 - GCD and LCM （质因式分解　GCD LCM）

最新推荐文章于 2021-07-18 23:58:10 发布

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#数论 #algorithm #组合数学

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本文探讨了在给定两个正整数G和L的情况下，如何计算满足特定条件的三元组(x, y, z)的数量，条件是它们的最大公约数等于G且最小公倍数等于L。

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 986 Accepted Submission(s): 459

Problem Description

Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying 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 two different solutions.

Input

First line comes an integer T (T <= 12), telling the number of test cases.
The next T lines, each contains two positive 32-bit signed integers, G and L.
It’s guaranteed that each answer will fit in a 32-bit signed integer.

Output

For each test case, print one line with the number of solutions satisfying the conditions above.

Sample Input

Sample Output

  
   72 
0

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

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题意：

输入G,L求有多少对数(x, y, z)满足gcd(x,y,z) = G lcm(x,y,z) = L

和上一个题类似

L/G质因数分解为p1^r1 * p2^r2 ... pn^rn

x/G分解为p1^a1 * p2^a2 ... pn^an

y/G z/G同理为 ...pi^bi... ...pi^ci...
x/G y/G z/G 这三个数肯定互质(GCD性质) 所以　min(ai, bi, ci) ==0

而lcm(x/G, y/G, z/G) = L/G 所以　max(ai, bi, ci) == ri

所以根据组合数学可得对于pi^ri这一项有

一个选择ri 一个选择0 另外一个随便选择然后全排列就是 6*(ri+1)　然后减去重复的6种（C(3,2)为0 +C(3,2)为ri）

即　6*(ri+1) - 6 种方案数

然后因为每一项相互独立，乘法原理相乘即可

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 70000 + 20;
int vis[maxn], prim[maxn];

void sieve(int n) {
    int m = (int) sqrt(n+0.5);
    memset(vis, 0, sizeof(vis));
    for(int i=2; i<=m; i++) if(!vis[i]) {
        for(int j=i*i; j<=n; j+=i) {
            vis[j] = 1;
        }
    }
}

int getPrimeTable(int n) {
    sieve(n);
    int c = 0;
    for(int i=2; i<=n; i++) if(!vis[i])
        prim[c++] = i;
    return c;
}

int main() {
    int T;

    int primNum = getPrimeTable(70000);
    scanf("%d", &T);
    while(T--) {
        int G, L;
        scanf("%d%d", &G, &L);
        if(L % G) {
            puts("0");
            continue;
        }
        LL ans = 1;
        int x = L / G;
        int m = (int) sqrt(x + 0.5);
        for(int i=0; i<primNum && prim[i] <= m && x > 1; i++) {
            int t = 0;
            while(x % prim[i] == 0) {
                t++;
                x /= prim[i];
            }
            if(t) {
                ans *= 6 * (t + 1) - 6;
            }
        }
        if(x > 1) ans *= 6;
        P64I(ans);
    }

    return 0;
}