这道题需要枚举。如果直接枚举会$TLE$。

考虑进制的转换：对于$> x$的进制下，一定是回文数

回文长度$2$位：设每一位为$i$，进制为$x$，则该数为$i*x+i$。反之，如果$n=i*(x+1)$，则$x$进制下$n$为回文。但要满足$i<x$，所以$x>\sqrt{n}$时适用。

当$x \leq \sqrt{n}$时暴力判断，这样复杂度降为$O(T \sqrt{n})$。

1 #include 
     
       2  3 using namespace std; 4  5 #define re register 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i) 7 #define repd(i, a, b) for (re int i = a; i >= b; --i) 8 #define maxx(a, b) a = max(a, b); 9 #define minn(a, b) a = min(a, b);10 #define LL long long11 #define INF (1 << 30)12 13 inline LL read() {14     LL w = 0; int f = 1; char c = getchar();15     while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();16     while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();17     return w * f;18 }19 20 int T, l[40];21 LL n;22 23 int main() {24     T = read();25 26     rep(kase, 1, T) {27         n = read();28         if (n == 2) printf("%d\n", 3);29         else if (n <= 3) printf("%d\n", 2);30         else {31             register LL ans = INF;32             for (register int i = sqrt(n-1); i; i--)33                 if (!(n - n / i * i)) {34                     ans = n / i - 1;35                     break;36                 }37             register int range = sqrt(n), len;38             register bool flag;39             register LL x;40             for (register int i = 2; i <= range; ++i) {41                 x = n;42                 len = 0;43                 while (x) {44                     l[++len] = x - x / i * i;45                     x /= i;46                 }47                 flag = 1;48                 for (register int j = 1; j <= len / 2; j++)49                     if (l[j] != l[len - j + 1]) {50                         flag = 0;51                         break;52                     }53                 if (flag) { ans = i; break; }54             }55             printf("%lld\n", ans);56         }57     }58 59     return 0;60 }
     

 另外这道题比较卡常，需要一定的优化。