zoj 3816 Generalized Palindromic Number (二分+贪心)

  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 }