1 //反向62 2 #include <iostream> 3 #include <algorithm> 4 #include <string> 5 #include <cstring> 6 #include <cstdio> 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 - ans + 1); 49 } 50 return 0; 51 }