uva 725 Division(暴力枚举) 解题心得

原题：

Description

 

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9once 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

 

 分析：虽然是采用暴力的方法，但是也要有技巧的暴力， 并不是无脑暴力，只要暴力前5个数的可能，就能根据n的值推断出后5个数

　　，再进行判断数字有没有重复就可以了，最后还要注意输出的格式，空位用‘0’填充

 

我的代码：

#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;
}