Given a natural number N, find an M such that M > 0 and M is a multiple of N, and the decimal representation of M contains only 0 or 1. Find the smallest M.
For example: N=4, M=100.
Input
Enter a number N. (1 <= N <= 10^6)
Output
Output the smallest M that meets the conditions.
Input example
4
Output example
100
data may be very large, so long long will burst, so the remainder can be handled, and the remainder cannot exceed 999999.
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <map> #include <set> using namespace std; #pragma comment(linker, "/stck:1024000000,1024000000") #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.1415926535897932384626433832 #define ios() ios::sync_with_stdio(true) #define INF 0x3f3f3f3f #define mem(a) ((a,0,sizeof(a))) int n; bool vis[1000006]; struct node { string str; int data; }years,pos; void dfs(int n) { queue<node>q; memset(vis,0,sizeof(vis)); ans.str="1"; ans.data = 1 % n; vis[ 1 %n]= 1 ; q.push(ans); while(!q.empty()) { pos=q.front(); q.pop(); if(pos.data%n==0) { cout<<pos.str<<endl; return ; } for(int i=0;i<2;i++) { if(!vis[(pos.data*10+i)%n]) { ans.str = pos.str; ans.str+=i+'0'; ans.data=(pos.data*10+i)%n; vis [ans.data] = 1 ; q.push(ans); } } } } intmain () { scanf("%d",&n); dfs(n); return 0; }