hdu 4497 GCD and LCM (唯一分解定理 + 计数)
2016-08-02 21:02
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GCD and LCM
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2262 Accepted Submission(s): 1000
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, print one line with the number of solutions satisfying the conditions above.
[align=left]Sample Input[/align]
2
6 72
7 33
[align=left]Sample Output[/align]
72
0
[align=left]Source[/align]
2013 ACM-ICPC吉林通化全国邀请赛——题目重现
大体题意:
告诉你G和L,求出有多少个(X,Y,Z)对组合 ,满足, 他们的最大公约数为G,最小公倍数是L。
思路:
比赛时没有做出来,借鉴的学长的博客~~,其实推理的差不多了,就是无法实现最后的一步,还是练的少啊= = !
如果L不能整除G的话,那么答案就是0。
否则可以把G,2G,3G,4G,,,,,L 都除以G,变成1,2,3,4,5,,,n。
所以n = L/G。
把n 质因数分解:
得到: p1^t1 + p2^t2 + p3 ^ t3 + .......
那么符合条件的x,y,z满足:
x = p1^i1 + p2 ^i2 + p3 ^ i3 + .....
y = p1^j1 + p2 ^j2 + p3 ^ j3 + .....
z = p1^k1 + p2 ^k2 + p3 ^ k3 + .....
首先x,y,z最大公约数是1的话,那么i1,j1,k1 至少一个为0,至少一个为t1,但又不能全是:
所以所有的组合情况为:
0 0 t1 --> 总共3种组合
0 t1 t1 --> 总共3种组合
t1 0 1~t1-1 总共t1 - 1种组合!
那么总组合为 6*t1 - 6 + 3+ 3 = 6*t1种组合!
那么用唯一分解定理分解n 即可! 枚举t!
n 最大是int 2^ 31 那么只需要枚举 2^16进行分解即可!
详细见代码:
#include<cstdio> #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<deque> #include<queue> #include<stack> #include<list> typedef long long ll; typedef unsigned long long llu; const int maxn = 1000 + 10; const int inf = 0x3f3f3f3f; const double pi = acos(-1.0); const double eps = 1e-8; using namespace std; int len = 1 << 16; int main(){ int T; scanf("%d",&T); while(T--){ int a,b; scanf("%d %d",&a,&b); if (b % a != 0){ printf("0\n"); continue; } ll ans = 1ll; int n = b/a; for (int i = 2; i <= len; ++i){ if (n % i == 0){ int sum = 0; while(n % i == 0)n /= i,++sum; ans *= sum*6; } } if (n > 1)ans *= 6; printf("%I64d\n",ans); } return 0; }
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