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。
1 #include<cstdio> 2 #include<cmath> 3 #include<algorithm> 4 #include<cstring> 5 #include<iostream> 6 #include<map> 7 #define ll long long 8 typedef std::pair<ll,int> info; 9 std::map<info,info>mp; 10 ll p[20],n; 11 ll read(){ 12 ll t=0,f=1;char ch=getchar(); 13 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();} 14 while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();} 15 return t*f; 16 } 17 info calc(int idx,info x){ 18 if (idx==1){ 19 if (x.first>=x.second) return info(1,0); 20 else return info(0,x.first); 21 } 22 if (mp.find(x)!=mp.end()) return mp[x]; 23 int t=x.first/p[idx-1]; 24 ll num=x.first-t*p[idx-1]; 25 info ans(0,0); 26 for (int i=t;i>0;i--){ 27 info res=calc(idx-1,info(num,std::max(i,x.second))); 28 ans.first+=res.first+1; 29 num=p[idx-1]+res.second-std::max(i,x.second); 30 } 31 info res=calc(idx-1,info(num,x.second)); 32 ans.first+=res.first;ans.second+=res.second; 33 return mp[x]=ans; 34 } 35 int main(){ 36 n=read(); 37 p[0]=1; 38 for (int i=1;i<=18;i++) p[i]=p[i-1]*10; 39 int len=0; 40 for (ll x=n;x;x/=10) len++; 41 if (n==0) puts("0"); 42 else printf("%lld\n",calc(len,info(n,0)).first); 43 return 0; 44 }
