LightOJ

Finding LCM

已知a, b, c的最小公倍数为L, 给你a,b,问你是否存在最小的c满足题意，不存在输出impossible

素数分解

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long LL;

int T;
LL A, B, L;
const int maxn = 1e6 + 10;

int a[maxn], b[maxn], l[maxn];

bool is_prime[maxn];
int prime[maxn];

void init() {
    for (int i = 0; i < maxn; i++) is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i < maxn; i++) {
        if (is_prime[i]) {
            prime[++prime[0]] = i;
            for (int j = i + i; j < maxn; j += i) {
                is_prime[j] = false;
            }
        }
    }
}

void Break(LL &x, int c[]) {
    for (int i = 1; i <= prime[0] && x != 1; i++) {
        while (x % prime[i] == 0) {
            x /= prime[i];
            c[prime[i]]++;
        }
    }
}


void solve() {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    memset(l, 0, sizeof(l));
    Break(A, a); Break(B, b), Break(L, l);
    bool flag = 1;
    LL ans = 1;
    for (int i = 2; i < maxn && flag; i++) {
        if (a[i] > l[i] || b[i] > l[i]) {
            flag = 0;
        } else if (a[i] < l[i] && b[i] < l[i]) {
            ans *= pow((LL)i, (LL)l[i]);
        }
    }
    ans *= L;
    if (!flag) {
        puts("impossible");
    } else {
        printf("%lld
", ans);
    }
}

int main() {
    init();
    scanf("%d", &T);
    for (int t = 1; t <= T; t++) {
        scanf("%lld%lld%lld", &A, &B, &L);
        printf("Case %d: ", t);
        solve();
    }

    return 0;
}