题目链接：点击打开链接

题意：

给定一个数n

找一个最大的数u使得u<n && u为回文。

枚举前面有多少位是一样的。然后分类讨论。啪啦啪啦

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 22;
int pie[N], piesize;
ll ans, x;

int z[N], a[N], asize;
ll mx[N];

bool ok(ll v) {
	int x, dep = 0;
	while (v > 0) {
		x = v % 10;
		if (dep == 0 || x != z[dep - 1])
			z[dep++] = x;
		v /= 10;
	}	
	int l = 0, r = dep - 1;
	while (l < r) {
		if (z[l] == z[r])
			++ l, -- r;
		else
			return false;
	}
	return true;
}

void up(ll g) {
	if (g > ans) {
		ans = g;
		for (int i = 0; ;++i) {
			g /= 10;
			mx[i] = g;
			if (g == 0)
				return ;
		}
	}
}

void dfs(int dep, ll g, int idy, int f) {
	if (dep == -1) {
		
		if (g <= x && ok(g)) {
			up(g);
		}
	} else {
		if (mx[dep] > g)
			return ;	
			
		if (f) {
			dfs(dep - 1, g * 10 + a[idy], idy, 1);
			if (idy - 1 >= 0)
				dfs(dep - 1, g * 10 + a[idy - 1], idy - 1, 1);
		} else {
			if (a[idy] < pie[dep])	{
				dfs(dep - 1, g * 10 + a[idy], idy, 1);
			} else if (a[idy] == pie[dep]) {
				dfs(dep - 1, g * 10 + a[idy], idy, 0);
			}
			if (idy - 1 >= 0) {
				if (a[idy - 1] < pie[dep])
					dfs(dep - 1, g * 10 + a[idy - 1], idy - 1, 1);
				else if (a[idy - 1] == pie[dep])
					dfs(dep - 1, g * 10 + a[idy - 1], idy - 1, 0);
			}
		}
	}
}

/*
void dfs(int dep, ll g, int idy, int f) {
	if (dep == -1) {
		if (g <= x && ok(g))
			ans = max(ans, g);		
	} else {
		dfs(dep - 1, g * 10 + a[idy], idy);
		if (idy - 1 >= 0)
			dfs(dep - 1, g * 10 + a[idy - 1], idy - 1);
	}
}
*/

void work() {
	ll y;
	int n, v;
	
	//x = 999915494587545487;
	scanf("%lld", &x);
	-- x;
	if (ok(x)) {
		printf("%lld\n", x);
		return ;
	}
	if (ok(x - 1)) {
		printf("%lld\n", x - 1);
		return ;
	}
	//
	y = x;
	piesize = 0;
	while (y > 0) {
		pie[piesize ++] = y % 10;
		y /= 10;
	}
	n = piesize;
	ans = 0;
	memset(mx, 0, sizeof mx);
	ll g, cnt;
	for (int i = n - 1; i >= 0; --i) {
		if (pie[i] == 0)
			continue;
			
		asize = 0;
		for (int j = n - 1; j > i; -- j) {
			if (asize == 0 || a[asize - 1] != pie[j])
				a[asize ++] = pie[j];	
		}
		if (asize == 0 || a[asize - 1] != pie[i] - 1)
			a[asize ++] = pie[i] - 1;
			
		g = 0;
		for (int j = n - 1; j > i; --j) 
			g = g * 10 + pie[j];
		g = g * 10 + pie[i] - 1;
		dfs(i - 1, g, asize - 1, 1);
		
		cnt = i - asize;
		ll gg = g;
		for (int j = 0; j <= cnt; ++j)
			g = g * 10 + 9;
		for (int j = asize - 2; j >= 0; --j)
			g = g * 10 + a[j];
		if (ok(g) && g <= x)
			up(g);
		
		for (int j = 0; j < cnt; ++j)
			gg = gg * 10 + 9;
		for (int j = asize - 1; j >= 0; --j)
			gg = gg * 10 + a[j];
		
		if (gg <= x && ok(gg))
			up(gg);
	}
	for (int i = n - 1; i >= 0; --i) {
		asize = 0;
		g = 0;
		for (int j = n - 1; j >= i; --j) {
			g = g * 10 + pie[j];
			if (asize == 0 || a[asize - 1] != pie[j])
				a[asize ++] = pie[j];
		}
		dfs(i - 1, g, asize - 1, 0);
	}
	printf("%lld\n", ans);
}

int main() {
	int cas;
	scanf("%d", &cas);
	while (cas -- > 0) 
		work();
	return 0;
}

ZOJ 3816 Generalized Palindromic Number dfs+暴力枚举