题目链接：zoj 3816 Generalized Palindromic Number

题目大意：给定n，找一个最大的数x，保证x小于n，并且x为palindromic number

解题思路：枚举前i个放于n相同的数，然后去构造后半部分即可。

#include 
   
     #include 
    
      #include 
     
       using namespace std; typedef unsigned long long ll; int n = 0, bit[30], num[30], ans[30]; bool judge(int* a, int* b, int n) { for (int i = n - 1; i >= 0; i--) { if (a[i] != b[i]) return a[i] > b[i]; } return false; } bool cmp (int* a, int* b, int d) { for (int i = 1; i <= d; i++) { if (a[n-i] != b[n-i]) return a[n-i] > b[n-i]; } return false; } void dfs (int d, int p, bool flag) { if (d < p) { if (judge(bit, num, n) && judge(num, ans, n)) memcpy(ans, num, sizeof(num)); return; } int end = (flag ? bit[d] : 9); for (int i = end; i >= 0; i--) { int v = p; if (cmp(ans, num, n - d - 1)) return; num[d] = i; if (num[d] != num[d+1]) { while (v <= d) { num[v++] = num[d]; dfs(d - 1, v, flag && i == end); } } else dfs(d - 1, p, flag && i == end); } } ll solve () { ll a; scanf("%llu", &a); memset(ans, 0, sizeof(ans)); memset(bit, 0, sizeof(bit)); memset(num, 0, sizeof(num)); n = 0; while (a) { bit[n++] = a % 10; a /= 10; } num[n] = 0; dfs(n-1, 0, 1); ll ret = 0; for (int i = n - 1; i >= 0; i--) ret = ret * 10 + ans[i]; return ret; } int main () { int cas; scanf("%d", &cas); while (cas--) { printf("%llu\n", solve()); } return 0; }