把全部合法的进制打出来会发现合法的进制都是在 

n/3 n/4 n/5的边上

然后暴力边上的进制数。。

#include 
     
      #include 
      
       typedef long long ll;bool ok(ll x, ll y) {    ll v;    while (x > 0) {        v = x % y;        if (v != 3 && v != 4 && v != 5 && v != 6)             return false;        x /= y;    }        return true;}std::set
       
         vis;int T = 0;ll n;void work() {    scanf("%I64d", &n);    printf("Case #%d: ", ++T);    if (n == 3 || n == 4 || n == 5 || n == 6)        puts("-1");    else {        vis.clear();        int ans = 0;        ll base;        for (int i = 1; i <= 50 && i < n; ++i) {            for (int j = 0; j <= 50; ++j) {                base = n / i + j;                if (base <= 1 || base >= n || vis.count(base) > 0)                    continue;                vis.insert(base);                if (ok(n, base))                    ++ ans;            }            for (int j = 1; j <= 50; ++j) {                base = n / i - j;                if (base <= 1 || base >= n || vis.count(base) > 0)                    continue;                vis.insert(base);                if (ok(n, base))                    ++ ans;            }        }        for (int i = 2; i < n && i <= 1000; ++i) {            if (vis.count(i))                continue;            vis.insert(i);            if (ok(n, i))                ++ ans;        }        printf("%d\n", ans);    }}int main() {    int cas;    scanf("%d", &cas);    while (cas -- > 0)         work();    return 0;}