[UVa] 725 - Division

以Brute-force去求解
輸入N，求 N = abcde / fghij
而a~j為0~9的數字，互相不重複
每兩個輸出間要空行，而最後以0結束時又不能多空行

#include <stdio.h>
#include <stdlib.h>
#include <string>
using namespace std;

int N,a,b; // a / b
int main(){
  scanf("%d",&N);
  while(1){
    int ok=0;
    if(N==0)break;
    else {
      for(int k=0;k<100000;k++){
        int b1 = k/10000,
        b2 = k%10000/1000,
        b3 = k%10000%1000/100,
        b4 = k%10000%1000%100/10,
        b5 = k%10000%1000%100%10;
        //printf("%d %d %d %d\n",b1,b2,b3,b4);
        if(b1!=b2 && b1!=b3 && b1!=b4 && b1!=b5 && b2!=b3 && b2!=b4 && b2!=b5
        && b3!=b4 && b3!=b5 && b4!=b5){
          b = k;
          //printf("%d ",b);
          a = b*N;
          if(a/10000 >=10)break;
          else {
            int a1 = a/10000,
            a2 = a%10000/1000,
            a3 = a%10000%1000/100,
            a4 = a%10000%1000%100/10,
            a5 = a%10000%1000%100%10;
            if(   a1!=a2 && a1!=a3 && a1!=a4 && a1!=a5
               && a1!=b1 && a1!=b2 && a1!=b3 && a1!=b4 && a1!=b5
               && a2!=a3 && a2!=a4 && a2!=a5
               && a2!=b1 && a2!=b2 && a2!=b3 && a2!=b4 && a2!=b5
               && a3!=a4 && a3!=a5
               && a3!=b1 && a3!=b2 && a3!=b3 && a3!=b4 && a3!=b5
               && a4!=a5 && a4!=b1 && a4!=b2 && a4!=b3 && a4!=b4 && a4!=b5
               && a5!=b1 && a5!=b2 && a5!=b3 && a5!=b4 && a5!=b5
              )
            {
              printf("%05d / %05d = %d\n",a,b,N); 
              ok=1;
            }
          }
        }
      }
        if(ok==0)printf("There are no solutions for %d.\n",N);
    }
    scanf("%d",&N);
    if(N==0)break; else puts("");

  }
    return 0;
}




Ruija Liu 的神奇寫法
將運算式轉為字串
按照大小排序後，直接檢查0~9是否有缺數字
即可作為題目限制的判定

// UVa725 Division
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main() {
  int n, kase = 0;
  char buf[99];
  while(scanf("%d", &n) == 1 && n) {
    int cnt = 0;
    if(kase++) printf("\n");
    for(int fghij = 1234; ; fghij++) {
      int abcde = fghij * n;
      sprintf(buf, "%05d%05d", abcde, fghij);
      if(strlen(buf) > 10) break;
      sort(buf, buf+10);
      bool ok = true;
      for(int i = 0; i < 10; i++)
        if(buf[i] != '0' + i) ok = false;
      if(ok) {
        cnt++;
        printf("%05d / %05d = %d\n", abcde, fghij, n);
      }
    }
    if(!cnt) printf("There are no solutions for %d.\n", n);
  }
  return 0;
}