【动态规划】数位dp

优雅的数字

只要一个正整数的十进制表示中包含不超过 3 个非零数字，就被认为是优雅的数字。例如，3、2000、123 是优雅的数字，而 4321、12306、120086 则不是。

#include<iostream>
using namespace std;
#define ll long long
#include<cstring>
ll a[20];
ll f[20][2][20];
/*
2
11111 22222
1 1000
*/
ll dfs(ll pos,bool st,bool limit,ll cnt){
if(pos<0)return (ll)(cnt<=3);
if(!limit&&f[pos][st][cnt]!=-1)return f[pos][st][cnt];//再写记忆化


ll res=0;
ll up=limit?a[pos]:9;
for(ll i=0;i<=up;i++){
    res+=dfs(pos-1,st||(i>0),limit&&(i==up),cnt+(ll)(i>0));
}

if(!limit)f[pos][st][cnt]=res;//再写记忆化
return res;
}

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

int main(){
ll T;cin>>T;
memset(f,-1,sizeof(f));
while(T--){
        ll l,r;
    cin>>l>>r;
    cout<<sum(r)-sum(l-1)<<endl;
}

}

P2657 [SCOI2009] windy 数

#include<iostream>
using namespace std;
#define ll long long
#include<cstring>
ll f[10][2][2][10];
ll a[10];

ll dfs(ll pos,bool st,bool limit,bool ok,ll lst){
if(pos<0)return (ll)(ok);
if(!limit&&f[pos][st][ok][lst]!=-1)return f[pos][st][ok][lst];
ll res=0;
ll up=limit?a[pos]:9;
for(ll i=0;i<=up;i++){
    res+=dfs(pos-1,st||(i>0),limit&&(i==up),ok&&(!st||abs(i-lst)>=2),i);
}
f[pos][st][ok][lst]=res;
return res;
}
ll sum(ll x){
ll pos=-1;
while(x){
    a[++pos]=x%10;
    x/=10;
}return dfs(pos,0,1,1,0);
}
int main(){
    memset(f,-1,sizeof(f));
    ll x,y;
cin>>x>>y;
cout<<sum(y)-sum(x-1);
}

7的意志

#include <iostream>
#include<cstring>
#define ll long long
using namespace std;
const ll N=20;
ll n,m;
ll a[N];
ll f[20][2][10][10];
ll  dfs(ll pos,bool st,bool limit,ll num,ll sum){
if(pos<0)return (ll)(num==0&&sum==0);
if(!limit&&f[pos][st][num][sum]!=-1)return f[pos][st][num][sum];
ll res=0;
ll up=limit?a[pos]:9;
for(ll i=0;i<=up;i++){
  res+=dfs(pos-1,st||(i>0),limit&&(i==up),
  (num*10+i)%7,(sum+i)%7);
}
if(!limit)f[pos][st][num][sum]=res;
return res;
}
ll get(ll x){
ll pos=-1;
while(x){
  a[++pos]=x%10;
  x/=10;
}return dfs(pos,0,1,0,0);
}
int main()
{
  // 请在此输入您的代码

  memset(f,-1,sizeof(f));
  while(1){
scanf("%lld%lld",&n,&m);
if(!n&&!m)break;
printf("%lld\n",get(m)-get(n-1));
  }

  return 0;
}