Zoj 3816 generalized palindromic number (Binary + greedy)

Question connection: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 5348

The questions of the Mudanjiang online competition were introduced during the competition, but they have never been called out. After the competition, they have also been adjusted for some time. Recently, the Code capabilities are worrying ~

There are many ways to do this. My approach is to lower the bottom of two points, that is, to determine whether there is a qualified number between low and n-1. Then greedy to construct a solution (intuition tells me that it is very troublesome to directly construct the maximum solution ).

The constructor is like this: first, let the first few digits of low and n be the same, because in any case, these digits remain unchanged. Then, if one of them is different, therefore, the lower bound is not required for the number of places to be placed in the following position. Simply greedy for the number of rows, just compare the size of the number of rows and the number of N, that is, the final problem becomes determined by the first few digits, and the total number of digits is also determined. To solve this problem, we can be greedy, but we need to enumerate where the middle part is, note that you can place 0 in the middle. For details, see the code.

  1 #include <cstdio>  2 #include <cstring>  3 #include <algorithm>  4 #include <iostream>  5 #include <time.h>  6   7 using namespace std;  8 typedef long long lld;  9 const int MAXN = 55; 10 const lld INF = 1LL<<62; 11 int bit[MAXN]; 12 int BIT[MAXN]; 13 lld n; 14 int m; 15 int arr[MAXN]; 16  17 lld calc(int a[], int cnt, int mid, int inx) { 18     if (2 * mid - cnt > inx) return INF; 19     lld ret = 0; 20     for (int i = m; i >= inx; i--) { 21         ret *= 10; ret += arr[i]; 22     } 23     for (int i = mid + 1; i <= cnt; i++) { 24         if (2 * mid - i <= 0 || a[i] != a[2 * mid - i]) return INF; 25     } 26     int ind = 2 * mid - cnt; 27     for (int i = inx - 1; i >= 1 && ind >= 1; i--) { 28         ret *= 10; 29         if (ind == 1) { 30             ret += a[ind]; 31         }else { 32             if (ind-1 == i || a[ind-1] <= a[ind]) { 33                 ret += a[ind-1]; ind--; 34             }else { 35                 ret += a[ind]; 36             } 37         } 38     } 39     return ret; 40 } 41  42 int isok(int inx) { 43     int cnt = 0; 44     int a[MAXN]; 45     for (int i = m; i >= inx; i--) { 46         if (arr[i] != 0) { 47             a[++ cnt] = arr[i]; 48             for (int j = i - 1; j >= inx; j--) { 49                 if (a[cnt] != arr[j]) a[++ cnt] = arr[j]; 50             } 51             break; 52         } 53     } 54     for (int i = cnt/2; i <= cnt; i++) { 55         lld ret = calc(a, cnt, i, inx); 56         if (ret <= n) return 1; 57     } 58     a[++ cnt] = 0; 59     return calc(a, cnt, cnt, inx) <= n; 60 } 61  62 int solve() { 63     for (int i = m, fg = 0; i >= 1; i--) { 64         if (bit[i] == BIT[i] && !fg) { 65             arr[i] = bit[i]; continue; 66         } 67         int Max = (fg >= 1 ? 9 : BIT[i]); 68         for (int j = Max; j >= bit[i] + 1; j--) { 69             arr[i] = j; 70             if (isok(i)) return 1; 71         } 72         arr[i] = bit[i]; 73         fg ++; 74     } 75     int a[MAXN], cnt = 1; 76     a[1] = arr[m]; 77     for (int i = m - 1; i >= 1; i--) { 78         if (a[cnt] != arr[i]) a[++ cnt] = arr[i]; 79     } 80     for (int i = 1; i <= cnt/2; i++) { 81         if (a[i] != a[cnt-i+1]) return 0; 82     } 83     return 1; 84 } 85  86 int check(lld x) { 87     int len = 0; 88     memset(bit, 0, sizeof(bit)); 89     memset(arr, 0, sizeof(arr)); 90     while (x) { 91         bit[++ len] = x % 10; 92         x /= 10; 93     } 94     return solve(); 95 } 96 int main() { 97     int T; 98     scanf("%d", &T); 99     for (int cas = 1; cas <= T; cas++) {100         cin >> n;101         int ret;102         if (n == 1) {103             printf("0\n"); continue;104         }105         n = n - 1;106         lld l = 1, r = n;107         m = 0;108         lld tmp = n;109         while (tmp) {110             BIT[++ m] = tmp % 10;111             tmp /= 10;112         }113         while (r >= l) {114             lld mid = r + l >> 1;115             if (check(mid)) {116                 l = mid + 1;117             }else {118                 r = mid - 1;119             }120         }121         cout << l - 1 << endl;122     }123     return 0;124 }

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Zoj 3816 generalized palindromic number (Binary + greedy)