Codeforces Find a Number 简单Bfs

A. Find a Number

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integers dd and ss. Find minimal positive integer nn which is divisible by dd and has sum of digits equal to ss.

Input

The first line contains two positive integers dd and ss (1≤d≤500,1≤s≤50001≤d≤500,1≤s≤5000) separated by space.

Output

Print the required number or -1 if it doesn't exist.

Examples

input

Copy

13 50

output

Copy

699998

input

Copy

61 2

output

Copy

1000000000000000000000000000001

input

Copy

15 50

output

Copy

-1

题目的意思就不多说了，就是求一个数n对d取余等于0，其各个位上的数字和等于s

简单的Bfs就过了。。。

题目链接：http://codeforces.com/problemset/problem/1070/A

 

#include <bits/stdc++.h>
using namespace std ;
const int Maxn = 5e3 + 10 ;

struct Node{
    int len ;
    int mod ;
    int sum ;
    char s[600] ;
    Node(){
        len = -1 ;
        mod = sum = 0 ;
        memset(s, 0, sizeof(s)) ;
    }
};

int Vis[510][Maxn] ;

int main (){
    int d, s ;
    cin >> d >> s ;
    queue < Node > que ;
    Node New ;
    que.push(New) ;
    while (!que.empty()){
        New = que.front() ;
        que.pop() ;
        for (int i = 0; i <= 9; i++){
            Node p = New ;
            p.len++ ;
            p.s[p.len] = i + '0' ;
            p.mod = (p.mod * 10 + i) % d ;
            p.sum += i ;
            if (Vis[p.mod][p.sum] || p.sum > s) continue ;
            Vis[p.mod][p.sum] = 1 ;
            if (p.mod == 0 && p.sum == s){
                puts(p.s) ;
                return 0 ;
            }
            que.push(p) ;
        }
    }
    puts("-1") ;
    return 0 ;
}