[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$O(T sqrt{n})$。

 1 #include <bits/stdc++.h>
 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 long
11 #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
", 3);
29         else if (n <= 3) printf("%d
", 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
", ans);
56         }
57     }
58 
59     return 0;
60 }

另外这道题比较卡常，需要一定的优化。