Codeforces 489C Given Length and Sum of Digits...

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 >= 0 && 9 * 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 = 0; i < m; ++i){
            for(d = 0; d < 10; ++d){
                if((i > 0 || d > 0 || 1 == m && 0 == d) && judge(m - i - 1, sum - d)){ //handle preamble 0
                    a.push_back(‘0‘ + 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 = 0; i < m; ++i){
            for(d = 9; d >= 0; --d){
                if(judge(m - i - 1, sum - d)){
                    b.push_back(‘0‘ + d);
                    sum -= d;
                    break;
                }
            }
        }

        for(i = 0; i < a.size(); ++i){
            printf("%c",a[i]);
        }
        printf("\n");
        for(i = 0; i < b.size(); ++i){
            printf("%c",b[i]);
        }
        printf("\n");
    }
    return 0;
}