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;

}