XJOI网上同步训练DAY5 T3

就是对于一个数，我们去考虑把t*****减到(t-1)9999*的代价。

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<map>
 #define ll long long
 typedef std::pair<ll,int> info;
 std::map<info,info>mp;
 ll p[],n;
 ll read(){
     ll t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 info calc(int idx,info x){
     if (idx==){
         if (x.first>=x.second) return info(,);
         else return info(,x.first);
     }
     if (mp.find(x)!=mp.end()) return mp[x];
     int t=x.first/p[idx-];
     ll num=x.first-t*p[idx-];
     info ans(,);
     for (int i=t;i>;i--){
         info res=calc(idx-,info(num,std::max(i,x.second)));
         ans.first+=res.first+;
         num=p[idx-]+res.second-std::max(i,x.second);
     }
     info res=calc(idx-,info(num,x.second));
     ans.first+=res.first;ans.second+=res.second;
     return mp[x]=ans;
 }
 int main(){
     n=read();
     p[]=;
     for (int i=;i<=;i++) p[i]=p[i-]*;
     int len=;
     for (ll x=n;x;x/=) len++;
     if (n==) puts("");
     else printf("%lld\n",calc(len,info(n,)).first);
     return ;
 }