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uva 725 Division（暴力枚举）

2024-11-15 11:50:01   203人阅读

                                uva 725  Division


Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits0 through 9 once each, such that the first number divided by the second is equal to an integerN, where 技术分享. 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 forN.". 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


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。

解题思路：暴力枚举所有情况，然后判断是否成立。



#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;
}




uva 725 Division（暴力枚举）


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