loj 1032 数位dp

题目链接：http://lightoj.com/volume_showproblem.php?problem=1032

思路：数位dp, 采用记忆化搜索， dp[pos][pre][have] 表示 pos处，前一位为pre, 当前有have个满足条件的状态。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std; long long dp[][][];
 int n, digit[]; long long dfs(int pos, int pre, int have, int doing)
 {
     if (pos == -) {
         return have;
     }
     if (!doing && dp[pos][pre][have] != -) {
         return dp[pos][pre][have];
     }
     int end = doing ? digit[pos] : ;
     long long ans = ;
     for (int i = ; i <= end; i++) {
         int nhave = have;
         if (pre ==  && i == ) {
             nhave++;
         }
         ans += dfs(pos - , i, nhave, i == end && doing);
     }
     if (!doing) {
         dp[pos][pre][have] = ans;
     }
     return ans;
 }     long long Solve(int n)
 {
     int pos = ;
     while (n) {
         digit[pos] = n % ;
         n /= ;
         pos++;
     }
     return dfs(pos - , , , );
 } int main()
 {
     memset(dp, -, sizeof(dp));
     int _case, t = ;
     scanf("%d", &_case);
     while (_case--) {
         scanf("%d", &n);
         printf("Case %d: %lld\n", t++, Solve(n));
     }
     return ;
 }