POJ 1426 Find The multiple idea, linear congruence, search difficulty: 2

http://poj.org/problem?id=1426

Tested, all values from 1-200 have a long long solution, so can be directly stored with a long long

Starting from 1, each transfer to 10*s and 10*s+1, only the remainder can be stored,

For the remainder I, only the first remainder of I is the most useful, only record this value can be

#include <cstdio> #include <cstring> #include <queue>using namespace Std;const int Maxn=222;typedef                Long Long ll;ll dp[maxn];int n;queue<int >que;int main () {while (scanf ("%d", &n) ==1&&n) {                if (n==1) {puts ("1"); continue;}               Memset (Dp,-1,sizeof (DP));               while (!que.empty ()) Que.pop ();               Dp[1]=1;               Que.push (1);               BOOL Fl=false;                        while (!que.empty ()) {int S=que.front (); Que.pop ();                                if (s==0) {printf ("%i64d\n", Dp[s]);                                Fl=true;                        Break                        } int t=s*10%n;                                if (dp[t]==-1) {dp[t]=10*dp[s];                        Que.push (t);                        } t= (s*10+1)%n;                   if (dp[t]==-1) {             dp[t]=10*dp[s]+1;                        Que.push (t);        }} if (!FL) puts ("-1"); } return 0;}

　　

POJ 1426 Find The multiple idea, linear congruence, search difficulty: 2