Fzu problem 1895 divisible 45 problem

This problem is a bit burning, but as long as you want to know the number divisible by 45, you can be divisible by 5 and 9, can be divisible by 9 of the number of people Plus is definitely 9 multiples, can be divisible by 5 The end is 0 or 5.

Then the DFS process is slightly less understood, and there are a few optimizations that must be noted. The process of Dfs is to choose which number we do not, and the less the number the better, so the number of deletions in the DFS process should be getting smaller, this step must have, otherwise time out.

The output of the time also need to pay attention to the situation of 0, can only output a 0, the following is the code

#include <stdio.h>#include<string.h>#include<algorithm>using namespacestd;Const intN =1005;Const intM = the;Const intdir[5]= {0,5};inttmp, Rec[m], cnt[m];intN, S[n], x, id;BOOLFlag;CharAns[n];BOOLcmpChar*a,Char*b) {     for(intI=0; A[i]; i++)    {        if(A[i]!=b[i])returnA[i] <B[i]; }    return false;}voidDfsintMintDintsum) {    if(d>=m) {if((sum-tmp)%9==0)        {            CharNow[n]; Memset (now,0,sizeof(now)); intLen =0;  for(intI=9; i>=0; i--)            {                 for(intj =0; J < Cnt[i]; J + +) Now[len+ +] = i +'0'; } Now[len+ +] = Dir[id] +'0'; if(M < x | | (CMP (ans, now) && m = =x)) {flag=true; X=m;            strcpy (ans, now); }        }        return; }     for(inti =1; I <=8; i++)    {        if(!cnt[i])Continue; Cnt[i]--; DFS (M, D+1, sum+i); Cnt[i]++; }}intMain () {intCAs; scanf ("%d", &CAs);  for(intI=1; i<=cas; i++)    {        CharNum[n]; scanf ("%s", num); X= strlen (num), TMP =0; Flag=false; memset (REC,0,sizeof(REC)); memset (ans,0,sizeof(ans));  for(intI=0; i<x; i++)        {            intcur = num[i]-'0'; TMP+=cur; Rec[cur]++; }         for(intI=0; i<2; i++) {memcpy (CNT, REC,sizeof(REC)); Cnt[dir[i]]--; if(Cnt[dir[i]) <0)Continue; ID=i;  for(intj=0; J<=tmp && j<=x; J + +) Dfs (J,0,0); }        if(!flag) {printf ("impossible\n"); Continue; }        intLenss = strlen (ans), sum1 =0;  for(inti =0; i < Lenss; i++) {sum1+ = Ans[i]-'0'; }        if(Sum1 = =0) {puts ("0"); Continue;    } puts (ans); }    return 0;}

Fzu problem 1895 divisible 45 problem