湖南大学第十四届ACM程序设计新生杯 F. Find the AFei Numbers(数位dp)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int a[20];
ll dp[20][10][10];

ll dfs(int pos, int p1, int p2, bool limit){
  if(pos == -1) return 1;
  if(!limit && dp[pos][p1][p2] != -1) return dp[pos][p1][p2];
  int up = limit ? a[pos] : 9;
  ll ans = 0;
  for(int i=0;i<=up;i++){
    if(p1 == 5 && p2 == 2 && i == 0) continue;
    ans += dfs(pos - 1, p2, i, limit && i == a[pos]);
  }
  if(!limit) dp[pos][p1][p2] = ans;
  return ans;
}

ll solve(ll x){
  int pos = 0;
  while(x){
    a[pos++] = x % 10;
    x /= 10;
  }
  return dfs(pos - 1, 0, 0, true);
}

int main()
{
  int T;
  scanf("%d",&T);
  memset(dp, -1, sizeof(dp));
  while(T--){
    ll n;
    scanf("%lld", &n);
    printf("%lld\n", n - solve(n) + 1);
  }
  return 0;
}