1 //反向62 2 #include 
     
       3 #include 
      
        4 #include 
       
         5 #include 
        
          6 #include 
         
           7 using namespace std; 8  9 typedef long long ll;10 // const int maxn = 1e5+5;11 ll n;12 ll a[100];13 ll dp[100][100];14 15 //下标，前面是否4，是否有前导0， 是否有限制16 ll dfs(ll pos, ll sta, ll pre, ll limit){17     if(pos == -1) return 1;18     if(!limit && dp[pos][sta] != -1)19         return dp[pos][sta];20     int up = limit?a[pos]:9;21     ll ans = 0;22     for(int i = 0;i <= up;i++){23         if(pre == 4 && i == 9){24             continue;25         }26         ans += dfs(pos-1, i == 4, i, limit && i == a[pos]);27     }28     if(!limit) dp[pos][sta] = ans;29     return ans;30 }31 32 ll solve(ll x){33     ll pos = 0;34     while(x){35         a[pos++] = x%10;36         x /= 10;37     }38     return dfs(pos-1, 0, -1, true);39 }40 41 int main(){42     memset(dp, -1, sizeof dp);43     int t;44     scanf("%d", &t);45     while(t--){46         scanf("%lld", &n);47         ll ans = solve(n);48         printf("%lld\n", n - ans + 1);49     }50     return 0;51 }