Description
You are given two positive integers ddd and sss. Find minimal positive integer nnn which is divisible by ddd and has sum of digits equal to sss.
Input
The first line contains two positive integers ddd and sss(1≤d≤500,1≤s≤5000)(1≤d≤500,1≤s≤5000)(1≤d≤500,1≤s≤5000) separated by space.
Output
Print the required number or -1 if it doesn’t exist.
Sample Input
Input
13 50
Output
699998
Input
61 2
Output
1000000000000000000000000000001
Input
15 50
Output
-1
这道题是看了别人后的代码才写的，看到代码后没想到居然是个BFS（果然还是自己太菜啊）
就是让你求一个数，这个数能被sss整除，且每位数相加=d=d=d。
看了别人ac的代码后发现其实很简单,运用同余定理就行了。。。。QAQ我怎么这么菜
思路
先根据位数进行BFS，如果位数和>s位数和>s位数和>s就不入队，剩下的每次入队前都modmodmod ddd就好了（控制数字大小别超intintint。然后当余数等于000（正好被ddd整除了），位数和等于sss的时候就是答案
代码如下
#include <queue>
#include <map>
#include <unordered_map>
#include <queue>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
#include <iostream>
#include <sstream>
#include <algorithm>
#define lowbit(a) (a&(-a))
#define _mid(a,b) ((a+b)/2)
#define _mem(a,b) memset(a,0,(b+3)<<2)
#define fori(a) for(int i=0;i<a;i++)
#define forj(a) for(int j=0;j<a;j++)
#define ifor(a) for(int i=1;i<=a;i++)
#define jfor(a) for(int j=1;j<=a;j++)
#define mem(a,b) memset(a,b,sizeof(a))
#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)
#define IO do{\ios::sync_with_stdio(false);\cin.tie(0);\cout.tie(0);}while(0)
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define debug(a) cout <<(a) << endl
using namespace std;struct node{int mod;int bit;string s;node(){};node(int m,int b,string ss){mod=m,bit=b,s=ss;}
};bool v[501][5001];
string bfs(int d,int s){queue<node>q;q.push(node(0,0,""));v[0][0] = true;while(!q.empty()){node buf = q.front();q.pop();if(buf.bit <= s){if(buf.mod == 0&&buf.bit==s)return buf.s;fori(10){int bmod = (buf.mod*10+i)%d;int bbit = buf.bit+i;if(!v[bmod][bbit]){v[bmod][bbit] = true;q.push(node(bmod,bbit,buf.s+(char)(i+'0')));}}}}return "-1";
}
int main() {int s,d;cin >> d>> s;cout << bfs(d,s) << endl;return 0;}
)
