Codeforces Round #277.5 (Div. 2) C. Given Length and Sum of Digits...

贪心

先构造最大的，将s拆成9的倍数和余数，前面排列9的倍数个9再加上一个余数，最后排0

然后把最大的放余数位置的数减去1，然后后面填上0，最后填上1，再做一次翻转，得到最小的

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>


using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;

#define CLR(x,y) memset(x,y,sizeof(x))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define eps 1e-9
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62

template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1; char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------

int main()
{
    int n,s;
    read(n),read(s);
    if(n==1 && !s){cout<<0<<' '<<0;return 0;}
    if(n*9<s || !s){cout<<-1<<' '<<-1;return 0;}
    string st;
    while(s>0)
    {
        st+=min(9,s)+'0';
        s-=min(9,s);
    }
    string st1=st;
    while(st1.size()!=n)
        st1+='0';
    if(st.size()!=n)
    {
        st[st.size()-1]--;
        for(int i=n-st.size(); i>=2; i--)
            st+='0';
        st+='1';
    }
    reverse(st.begin(),st.end());
    cout<<st<<' '<<st1;
}