传送门：点击打开链接
题意：给n个数字，要求用这n个数字去组成ip，要求每个数字都出现过，且ip里面去掉.后是一个回文
思路：大暴力，枚举的时候把ip分一半枚举就行，求出没有.的所有组合，然后再考虑添加.
#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1const int MX = 1e2 + 5;int A[MX], vis[MX], n;
int S[MX], pos[MX], w[MX], len;
char line[MX];
vector<string>ans;void solve() {int s1, s2, s3, s4;for(int a = 1; a <= 3; a++) {s1 = 0;for(int i = 1; i <= a; i++) s1 = s1 * 10 + S[i];if(s1 > 255 || (a > 1 && !S[1])) continue;for(int b = 1; b <= 3; b++) {s2 = 0;for(int i = 1; i <= b; i++) s2 = s2 * 10 + S[a + i];if(s2 > 255 || (b > 1 && !S[a + 1])) continue;for(int c = 1; c <= 3; c++) {s3 = 0;for(int i = 1; i <= c; i++) s3 = s3 * 10 + S[a + b + i];if(s3 > 255 || a + b + c >= len || (c > 1 && !S[a + b + 1])) continue;s4 = 0;for(int i = 1; i <= len - a - b - c; i++) s4 = s4 * 10 + S[a + b + c + i];if(s4 > 255 || (len - a - b - c > 1 && !S[a + b + c + 1])) continue;sprintf(line, "%d.%d.%d.%d", s1, s2, s3, s4);ans.push_back(line);}}}
}void DFS1(int p) {if(p > (len + 1) / 2) {int cnt = 0;for(int i = 1; i <= n; i++) cnt += vis[i] > 0;if(cnt != n) return;solve();return;}for(int i = 1; i <= n; i++) {vis[i]++;S[p] = S[len - p + 1] = A[i];DFS1(p + 1);vis[i]--;}
}int main() {//FIN;while(~scanf("%d", &n)) {ans.clear();memset(vis, 0, sizeof(vis));for(int i = 1; i <= n; i++) {scanf("%d", &A[i]);}for(len = 4; len <= 12; len++) {DFS1(1);}sort(ans.begin(), ans.end());printf("%d\n", (int)ans.size());for(int i = 0; i < (int)ans.size(); i++) {printf("%s\n", ans[i].c_str());}}return 0;
}