B. Quasi Binary

// Problem: B. Quasi Binary
// Contest: Codeforces - Codeforces Round #300
// URL: https://codeforces.com/problemset/problem/538/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-20 15:01:24
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
vector<int> alls, ans;
int cnt = 0x3f3f3f3f;
int n;
bool check(int x) {
    while (x) {
        for (int i = 2; i <= 9; i ++)
            if (x % 10 == i) return 0;
        x /= 10;
    }
    return 1;
}
void dfs (int u, int n, vector<int> s) {
    if (s.size() >= cnt) {
        return;
    }
    if (n <= 0) {
        if (n == 0) {
            cnt = s.size();
            ans = s;
        }
        return;
    }
    
    if (u - 1 >= 0) {
        dfs(u - 1, n, s);
        if (n >= alls[u]) {
            s.pb(alls[u]);
            dfs(u, n - alls[u], s);
            s.pop_back();
            
            s.pb(alls[u]);
            dfs(u - 1, n - alls[u], s);
            s.pop_back();
        }
        else {
            dfs(u - 1, n, s);
        }
    }
    else {
        if (n >= alls[u]) {
            s.pb(alls[u]);
            dfs(u, n - alls[u], s);
            s.pop_back();
        }
    }
}
void solve() {
    cin >> n;
    int num[100] = {0};
    int c = 1;
    while (n) {
        int r=  n % 10;
        for (int i = 0; i <= 10&& r;i  ++) {
            num[i] += c ;
            r --;
        }
        n /= 10;
        c *= 10;
    }
    int cnt = 0;
    for (int i = 0; i <= 10; i ++) {
        if (num[i])
            cnt ++;
    }
    cout << cnt << endl;
    for (int i = 0; i <= cnt; i ++) {
        if (num[i])
            cout << num[i] << ' ';
    }
    puts("");
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意给定一个数问最少用多少个01表示的十进制数

题解:我一开始有在想从大到小贪心。但是这个贪心是错误的，无法得到最小的。我们通过对每一位进行考虑。
例如各位是x，那么就有x个1，十位同理，只不过要累加。