[UVA] 725 - Division＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-05-15 22:17:18| 人2,852| 回1 | 上一篇 | 下一篇

[UVA] 725 - Division

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Division

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

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

得修改加速了, 因此有很糟的速度

#include <stdio.h>

int main() {
    int n, i, j, first = 1;
    while(scanf("%d", &n) == 1 && n) {
        if(!first)
            puts("");
        first = 0;
        int flag = 0;
        for(i = 1234; i <= 99999; i++) {
            if(i*n > 99999)
                break;
            char used[10] = {}, str[10], flag1 = 0;
            sprintf(str, "%05d", i);
            for(j = 0; str[j]; j++)
                used[str[j]-'0']++;
          &nsp; sprintf(str, "%05d", i*n);
            for(j = 0; str[j]; j++)
                used[str[j]-'0']++;
            for(j = 0; j < 10; j++)
                if(used[j] != 1)
                    flag1 = 1;
            if(!flag1) {
                flag = 1;
                printf("%05d / %05d = %d\n", i*n, i, n);
            }
        }
        if(!flag)
            printf("There are no solutions for %d.\n", n);
    }
    return 0;
}