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.
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
解析：通过枚举除数fghij可以算出abcde，然后将这几个数字字符排序一下查找是否所有数字都不相同。特别注意的是：当ab
cde+fghij超过十位时可以终止枚举。
代码如下：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 7 int main() 8 { 9 int n, k= 0; 10 char buf[99]; 11 while(scanf("%d", &n) == 1 && n) 12 { 13 14 int cnt = 0; 15 if(k++) 16 printf(" "); 17 for(int fghij = 1234; ; fghij++) 18 { 19 int abcde = fghij * n; //通过枚举fghij的值确定abcde的值 20 sprintf(buf, "%05d%05d", abcde, fghij); //将这两个字符格式化 21 if(strlen(buf) > 10) // 循环终止条件：abcde+fghij大于十位 22 break; 23 sort(buf, buf+10); //排序后，方便剔除有重复数字的buf。 24 bool ok = true; 25 for(int i = 0; i < 10; i++) 26 if(buf[i] != '0' + i) 27 ok = false; 28 if(ok) 29 { 30 cnt++; 31 printf("%05d / %05d = %d ", abcde, fghij, n); //输出格式要格外注意！！ 32 } 33 } 34 if(!cnt) 35 printf("There are no solutions for %d. ", n); 36 } 37 return 0; 38 }
注意暴力解题时要认真分析，避免枚举过多超内存。