Finding LCM

已知a, b, c的最小公倍數為L, 給你a,b,問你是否存在最小的c滿足題意，不存在輸出impossible

素數分解

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #define INF 0x3f3f3f3f
 6 #define MOD 1000000007
 7 using namespace std;
 8 typedef long long LL;
 9 
10 int T;
11 LL A, B, L;
 
12 const int maxn = 1e6 + 10;
13 
14 int a[maxn], b[maxn], l[maxn];
15 
16 bool is_prime[maxn];
17 int prime[maxn];
18 
19 void init() {
20     for (int i = 0; i < maxn; i++) is_prime[i] = true;
21     is_prime[0] = is_prime[1] = false;
22     for (int i = 2; i < maxn; i++) {
23         if (is_prime[i]) {
 
24             prime[++prime[0]] = i;
25             for (int j = i + i; j < maxn; j += i) {
26                 is_prime[j] = false;
27             }
28         }
29     }
30 }
31 
32 void Break(LL &x, int c[]) {
33     for (int i = 1; i <= prime[0] && x != 1; i++) {
34         while (x % prime[i] == 0
 
) {
35             x /= prime[i];
36             c[prime[i]]++;
37         }
38     }
39 }
40 
41 
42 void solve() {
43     memset(a, 0, sizeof(a));
44     memset(b, 0, sizeof(b));
45     memset(l, 0, sizeof(l));
46     Break(A, a); Break(B, b), Break(L, l);
47     bool flag = 1;
48     LL ans = 1;
49     for (int i = 2; i < maxn && flag; i++) {
50         if (a[i] > l[i] || b[i] > l[i]) {
51             flag = 0;
52         } else if (a[i] < l[i] && b[i] < l[i]) {
53             ans *= pow((LL)i, (LL)l[i]);
54         }
55     }
56     ans *= L;
57     if (!flag) {
58         puts("impossible");
59     } else {
60         printf("%lld\n", ans);
61     }
62 }
63 
64 int main() {
65     init();
66     scanf("%d", &T);
67     for (int t = 1; t <= T; t++) {
68         scanf("%lld%lld%lld", &A, &B, &L);
69         printf("Case %d: ", t);
70         solve();
71     }
72 
73     return 0;
74 }

LightOJ - 1215 Finding LCM