UVA 725 Division

0. Do not use recursion to construct a five-digit number, the direct for loop to the maximum is good, can be slightly pruning a loss, because the minimum number is 01234 from 1234, because the multiplier n is a minimum of 2 and the numerator is a maximum of five digits, so the denominator should be less than 50,000. So the for 1234 to 50000-1 is OK. Oh, there's another reason, fenmu*n>= 100,000 can be thrown away.

1. There's one more thing to repeat .

 while(i) {num[i%Ten]=1; I/=Ten; }     while(j) {Num[j%Ten]=1; J/=Ten; }     for(intI=1;i<Ten; i++) num[0]+=Num[i]; if(num[0]==Ten) {printf ("%d/%05d =%d\n", Fenzi,fenmu,n); OK=true; }

2. Don't forget that when the molecule is less than 10,000, remember that num[0]=1 0 is being used.

3.!!! scanf control format when really really really very useful%nd (n is a number) represents the output%d that number and then less than n bits from the left to start with a space to complement the%0ND represents the output%d that number and then less than n bit from the left to start with 0 complement

1#include <cstdio>2#include <cstring>3 intN;4 BOOLOK;5 intnum[ A];6 voidSolveintIintj)7 {8     intFenmu=i,fenzi=J;9     if(i<10000) num[0]=1;Ten      while(i) One     { Anum[i%Ten]=1; -I/=Ten; -     } the      while(j) -     { -num[j%Ten]=1; -J/=Ten; +     } -      for(intI=1;i<Ten; i++) +num[0]+=Num[i]; A     if(num[0]==Ten) at     { -printf"%d/%05d =%d\n", fenzi,fenmu,n); -ok=true; -     } - } - intMain () in { -     intflag=0; to      while(~SCANF ("%d",&N)) +     { -         if(n==0) Break; the         if(flag) printf ("\ n"); *flag=1; $ Panax Notoginsengok=false; -          for(intI=1234;i<50000; i++) the         { +memset (NUM,0,sizeof(num)); A             intj=i*N; the             if(j>=100000)Continue; + solve (i,j); -         } $         if(!ok) printf ("There is no solutions for%d.\n", n); $     } -     return 0; -}

UVA 725 Division