UVA 725，uva725

UVA 725，uva725

Description

 

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where . That is,

 

abcde / fghij =N

where each letter represents a different digit. The first digit of one of the numerals is allowed to be zero.

 

Input 

Each line of the input file consists of a valid integer N. An input of zero is to terminate the program.

 

Output 

Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator).

Your output should be in the following general form:

 

xxxxx / xxxxx =N

xxxxx / xxxxx =N

.

.

 

In case there are no pairs of numerals satisfying the condition, you must write ``There are no solutions for N.". Separate the output for two different values of N by a blank line.

 

Sample Input 

61
62
0

 

Sample Output 

There are no solutions for 61.

79546 / 01283 = 62
94736 / 01528 = 62

 

 

 

題意：輸入一個數n，從小到大輸出它的abcde / fghij = n的類型的式子（注意空格，可以有前導0）a-j是一個0到9的排列。如果沒有則輸出There are no solutions for 61.   格式要求：每測試一組案例，換一空行。（這輸出格式也是RLGL）

 

 

解題思路：枚舉，但是是從被除數枚舉，通過相乘算出除數。然後就只需要判斷除數是否大於100000，a-j是否相等。

 

 

代碼如下：(本來想自己寫的，就是自己不會用標記，最後還是沒有堅持下來，還是看了別人的博客，然後仿寫了。)

 

想問個問題,用 memse(m,0,sizeof(m)) 清零數組m，為什麼一定是sizeof(m),  不可以用m的長度10，因為當我用10的時候就連案例都通不過了....

求告知，沒錯我就是這麼菜.....(⊙﹏⊙)b

 

 

 

 

 1 #include <stdio.h>
 2 #include <cstring>
 3 int m[10];
 4 int panduan(int a,int b)
 5 {
 6     if(a>100000)
 7         return 0;
 8     memset(m,0,sizeof(m));
 9     //for(int i=0;i<10;i++)
10       //  m[i]=0;
11     if(b<10000)
12         m[0];       //容易忘記13     while(a)
14     {
15         m[a%10]=1;  //每一位都標記
16         a=a/10;     //標記了就除掉一位
17     }
18     while(b)
19     {
20         m[b%10]=1;
21         b=b/10;
22     }
23     int sum=0;
24     for(int j=0; j<10; j++)
25         sum+=m[j];
26     return sum==10;    //當sum等於10才返回
27 }
28 int main()
29 {
30     int n,k=0;
31     while(scanf("%d",&n)==1&&n)
32     {
33         if(k>0) printf("\n"); k++;   //輸出格要求
34         int flag=1;
35         for(int i=1234; i<1000000; i++)
36         {
37             if(panduan(n*i,i))
38             {
39                 printf("%d / %05d = %d\n",i*n,i,n);
40                 flag=0;
41             }
42         }
43         if(flag)
44             printf("There are no solutions for %d.\n",n);
45 
46     }
47     return 0;
48 }