HDU 2089 不要62（数位DP）

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中文题

思路：数位DP，dp[i][j]表示到第i位，最后一位为j的情况，然后根据数位DP进行转移即可

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 12;int n, m, d[N], dn, dp[N][N];void get(int x) {if (x == 0) {dn = 1;d[dn] = 0;return;}dn = 0;while (x) {d[++dn] = x % 10;x /= 10; } for (int i = 1; i <= dn / 2; i++) swap(d[i], d[dn - i + 1]);}int solve(int x) {get(x);int pre = 11, flag = 0;dp[0][11] = 1;for (int i = 1; i <= dn; i++) {for (int j = 0; j <= 9; j++) {dp[i][j] = 0;if (j == 4) continue;for (int k = 0; k <= 9; k++) {if (k == 4 || (k == 6 && j == 2)) continue;dp[i][j] += dp[i - 1][k];   }  }if (flag) continue;  for (int j = 0; j < d[i]; j++) {  if (j == 4 || (pre == 6 && j == 2)) continue;  dp[i][j]++;  }  if (d[i] == 4 || (pre == 6 && d[i] == 2)) flag = 1;  pre = d[i]; } int ans = 0; for (int i = 0; i <= 9; i++) ans += dp[dn][i];if (!flag) ans++;return ans;}int main() {while (~scanf("%d%d", &n, &m) && n || m) {printf("%d\n", solve(m) - solve(n - 1));}return 0;}

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