A - Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Su
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
Miguel Revilla
2000-08-31
程序分析：此题的意思就是输入正整数N，按从小到大的顺序输出所有形如abcde/ fghij=N的表达式，其中a~j恰好为数字0-9的一个排列（可以有前导0），2<=N<=79.在这里我们需要使用暴力法，枚举0-9
的所有排列？没这个必要，也会超时。所以只需要枚举fghij就可以算出abcde，然后判断是否所有数字都不相同即可。不仅程序简单，而且枚举量也从10！=3628800降低至不到1W而且当abcde和fghij加起来超过10位即可以终止枚举
程序代码：
#include<stdio.h>
int number[15];
int check(int a, int b) {
if (a > 98765) return 0;
for (int i = 0; i < 10; i++) {
number[i] = 0;
}
if (b < 10000) number[0] = 1;
while (a) {
number[a % 10] = 1;
a /= 10;
}
while ( b ) {
number[b % 10] = 1;
b /= 10;
}
int sum = 0;
for (int i = 0; i < 10; i++)
sum += number[i];
return sum == 10;
}
int main() {
int ans, cnt = 0;
while (scanf("%d", &ans) == 1, ans) {
if (cnt++) printf("\n");
int flag = 0;
for (int i = 1234; i < 99999; i++) {
if (check(i * ans, i)) {
printf("%05d / %05d = %d\n", i * ans, i, ans);
flag = 1;
}
}
if (!flag) {
printf("There are no solutions for %d.\n",ans);
}
}
return 0;
}