poj 1465 & zoj 1136 Multiple (BFS+余数重判)

Multiple

Description

a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digits X1,X2..XM (at least one), finds the smallest strictly positive multiple of N that has no other digits besides X1,X2..XM (if such a multiple exists).

Input

The input has several data sets separated by an empty line, each data set having the following format:

On the first line - the number N 
On the second line - the number M 
On the following M lines - the digits X1,X2..XM.

Output

For each data set, the program should write to standard output on a single line the multiple, if such a multiple exists, and 0 otherwise.

An example of input and output:

Sample Input

22
3
7
0
1

2
1
1

Sample Output

110
0

知识点：用到一个余数定理吧。 假设 x%n = e， y%n = e，那么 （x*10+k）%n = （e*10+k）%n = （y*10+k）%n。
      题目意思：给你m个数字（0~9），判断他们能否组成n的最小倍数。

解题思路：因为要是最小的，所以我们先对这m个数就行排序，那样得到的倍数肯定是最小的，还有就是，最重要的一点就是用到余数重判，若余数之前出现过，就不需要再判断了，这里进行了剪枝。还有就是poj要自己写queue而zoj则可以用STL中的queue。

贴出代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Node
{
    int pre;     //前一个数的位置
    int remainder;    //取得的余数
    int digit;   //当前这个数的值
} node[5005];

int num[15], mark[5005];

int cmp(const void * a, const void * b)
{
    return *(int *)a - *(int *)b;
}

void Output(int rear)   //将所求得的数输出
{
    if(node[rear].pre != -1)
    {
        Output(node[rear].pre);
        printf("%d", node[rear].digit);
    }
}

void BFS(int n,int m)
{
    struct Node temp;      //进行初始化操作
    int front = 0, rear = 1, Rem;
    memset(mark, 0, sizeof(mark));
    node[0].pre = -1;
    node[0].remainder = 0;

    while(front < rear)
    {
        for(int i = 0; i<m; i++)
        {
            Rem = (node[front].remainder*10+num[i])%n;     //取余
            if(!mark[Rem] && (node[front].pre!=-1 || num[i]>0))    //确保第一个数字不为0，和余数重判剪枝
            {
                mark[Rem] = 1;
                temp.pre = front;
                temp.remainder = Rem;
                temp.digit = num[i];
                node[rear++] = temp;
                if(Rem == 0)     //找到了最小的倍数
                {
                    Output(rear-1);
                    printf("\n");
                    return ;
                }
            }
        }
        front++;
    }
    printf("0\n");
}

int main()
{
    int n, m;
    while(scanf("%d", &n)!=EOF)
    {
        scanf("%d", &m);
        for(int i = 0; i<m; i++)
        {
            scanf("%d", &num[i]);
        }
        qsort(num, m, sizeof(num[0]), cmp);   //对数组进行升序排序，以确保是最小的倍数

        if(n == 0)
            printf("0\n");
        else
            BFS(n, m);
    }

    return 0;
}

poj 1465 & zoj 1136 Multiple (BFS+余数重判)