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.
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.
61 62 0
There are no solutions for 61. 79546 / 01283 = 62 94736 / 01528 = 62
题目大意:0~9十个数组成两个5位数(或0开头的四位数),要求两数之商等于输入的数据N。
#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;
}
原文地址:http://blog.csdn.net/llx523113241/article/details/43084717