[洛谷P2425]小红帽的回文数

原题传送门

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$O(T \sqrt{n})$。

 #include <bits/stdc++.h>

 using namespace std;

 #define re register
 #define rep(i, a, b) for (re int i = a; i <= b; ++i)
 #define repd(i, a, b) for (re int i = a; i >= b; --i)
 #define maxx(a, b) a = max(a, b);
 #define minn(a, b) a = min(a, b);
 #define LL long long
 #define INF (1 << 30)

 inline LL read() {
     LL w = ; int f = ; char c = getchar();
     while (!isdigit(c)) f = c == '-' ? - : f, c = getchar();
     while (isdigit(c)) w = (w << ) + (w << ) + (c ^ ''), c = getchar();
     return w * f;
 }

 int T, l[];
 LL n;

 int main() {
     T = read();

     rep(kase, , T) {
         n = read();
         if (n == ) printf("%d\n", );
         else if (n <= ) printf("%d\n", );
         else {
             register LL ans = INF;
             for (register int i = sqrt(n-); i; i--)
                 if (!(n - n / i * i)) {
                     ans = n / i - ;
                     break;
                 }
             register int range = sqrt(n), len;
             register bool flag;
             register LL x;
             for (register int i = ; i <= range; ++i) {
                 x = n;
                 len = ;
                 while (x) {
                     l[++len] = x - x / i * i;
                     x /= i;
                 }
                 flag = ;
                 for (register int j = ; j <= len / ; j++)
                     if (l[j] != l[len - j + ]) {
                         flag = ;
                         break;
                     }
                 if (flag) { ans = i; break; }
             }
             printf("%lld\n", ans);
         }
     }

     return ;
 }

另外这道题比较卡常，需要一定的优化。