m位长度，S为各位的和
利用贪心的思想逐位判断过去即可
详细的注释已经在代码里啦~
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define ll long long
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x))) const int INF = 0x3f3f3f3f; vector <char> a, b; bool judge(int m, int s){ //judge whether m long s sum valid
return s >= && * m >= s;
} int main(){
int i, j, d, m, s;
while(EOF != scanf("%d%d",&m,&s)){
if(!judge(m, s)){
printf("-1 -1\n");
continue;
}
a.clear();
b.clear(); int sum = s;
for(i = ; i < m; ++i){
for(d = ; d < ; ++d){
if((i > || d > || == m && == d) && judge(m - i - , sum - d)){ //handle preamble 0
a.push_back('' + d);
sum -= d;
break;
}
}
} if(a.size() != m){ // if exist an answer, it proves that both existing a, b
printf("-1 -1\n");
continue;
} sum = s;
for(i = ; i < m; ++i){
for(d = ; d >= ; --d){
if(judge(m - i - , sum - d)){
b.push_back('' + d);
sum -= d;
break;
}
}
} for(i = ; i < a.size(); ++i){
printf("%c",a[i]);
}
printf("\n");
for(i = ; i < b.size(); ++i){
printf("%c",b[i]);
}
printf("\n");
}
return ;
}