HDU 4937 lucky number (number theory)

HDU 4937 lucky number

Question Link

Assume that a number is given and X is given. Each hexadecimal digit has only 3, 4, 5, and 6. determine the number of X types. If there are infinite types of output-1

Idea: first, the 3, 4, 5, and 6 characters are unlimitedly determined, and it is easy to think that it is not proved. Then, the last digit of the enumerated number is 3, 4, 5, 6, then the hexadecimal number must be a factor from this number, because the remaining number must be a1x ^ 1 + a2x ^ 2 + a3x ^ 3... in this way, you only need to find the base in the factor to determine. The method for finding the factor is decomposed first and then DFS is searched. The time-out is returned when the direct try-out is performed.

Code:

#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <algorithm>using namespace std;typedef long long ll;const ll N = 1e6 + 5;int t, vis[N], pn = 0, cnt[N], fn;ll n, prime[N], fra[N];set<ll> ans;void getFra(ll n) {    fn = 0;    for (int i = 0; i < pn && n >= prime[i]; i++) {if (n % prime[i] == 0) {    fra[fn] = prime[i];    cnt[fn] = 0;    while (n % prime[i] == 0) {n /= prime[i];cnt[fn]++;    }    fn++;}    }    if (n != 1) {fra[fn] = n;cnt[fn++] = 1;    }}bool check(ll b) {    ll tmp = n;    while (tmp) {if (tmp % b < 3 || tmp % b > 6)    return false;tmp /= b;    }    return true;}void dfs(int now, ll sum) {    if (now == fn) {if (check(sum))    ans.insert(sum);return;    }    ll tmp = 1;    for (int i = 0; i <= cnt[now]; i++) {dfs(now + 1, sum * tmp);tmp *= fra[now];    }}void solve(ll n, ll bas) {    getFra(n);    dfs(0, 1);}bool judge(ll n) {    if (n == 3 || n == 4 || n == 5|| n == 6)return true;    return false;}int main() {    for (ll i = 2; i < N; i++) {if (vis[i]) continue;prime[pn++] = i;for (ll j = i * i; j < N; j += i)    vis[j] = 1;    }    int cas = 0;    scanf("%d", &t);    while (t--) {scanf("%I64d", &n);if (judge(n)) {    printf("Case #%d: -1\n", ++cas);    continue;}ans.clear();for (ll i = 3; i <= 6; i++)    solve(n - i, i);int out = ans.size();printf("Case #%d: %d\n", ++cas, out);    }    return 0;}