[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 = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

int T, l[40];
LL n;

int main() {
    T = read();

    rep(kase, 1, T) {
        n = read();
        if (n == 2) printf("%d
", 3);
        else if (n <= 3) printf("%d
", 2);
        else {
            register LL ans = INF;
            for (register int i = sqrt(n-1); i; i--)
                if (!(n - n / i * i)) {
                    ans = n / i - 1;
                    break;
                }
            register int range = sqrt(n), len;
            register bool flag;
            register LL x;
            for (register int i = 2; i <= range; ++i) {
                x = n;
                len = 0;
                while (x) {
                    l[++len] = x - x / i * i;
                    x /= i;
                }
                flag = 1;
                for (register int j = 1; j <= len / 2; j++)
                    if (l[j] != l[len - j + 1]) {
                        flag = 0;
                        break;
                    }
                if (flag) { ans = i; break; }
            }
            printf("%lld
", ans);
        }
    }

    return 0;
}

 另外这道题比较卡常，需要一定的优化。