http://poj.org/problem?id=1465
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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 Source |
题意：
给出一个整数N，和M个0～9的数，求N的一个最小倍数，且该数仅由这M个数构成，不存在则输出0。
分析：
如果存在最终的数，一定可以写成A1*10^(k-1)+A2*10^(k-2)+...+Ak,Ai属于给出的M个数的集合。k有可能很大，64位整数也可能存不下。注意到最后的结果是N的倍数，假设结果是X，则有X%N=0,注意到结果的多项式形式，显然能想到使用同余定理。我们需要从1位扩展到k位（当然要一步步来），可以用BFS，用余数来记录状态（只需要第一个），这样状态不超过N个，一旦余数为0,我们需要的结果就出来了。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn
using namespace std;
int X[10];
int n,m;
int q[5555];
int st[5555];
bool __hash[5555];
struct __node
{
int x,mod,fir;
}node[5555];
void write(int x)
{
int top=-1;
for (;~x;x=node[x].fir)
st[++top]=node[x].x;
while (top>=0)
printf("%d",st[top--]);
puts("");
}
void bfs()
{
int f=0,r=-1,cnt=0;
if (!n)
{
printf("%d\n",0);
return ;
}
memset(__hash,0,sizeof __hash);
for (int i=0;i<m;i++)
{
if (!X[i]) continue;
int mod=X[i]%n;
if (!mod)
{
printf("%d\n",X[i]);
return ;
}
if (__hash[mod]) continue;
__hash[mod]=true;
node[cnt]=(__node){X[i],mod,-1};
q[++r]=cnt;
cnt++;
}
while (f<=r)
{
int x=q[f++];
for (int i=0;i<m;i++)
{
int mod=(node[x].mod*10+X[i])%n;
if (__hash[mod]) continue;
__hash[mod]=true;
node[cnt]=(__node){X[i],mod,x};
q[++r]=cnt;
if (!mod)
{
write(cnt);
return ;
}
cnt++;
}
}
printf("0\n");
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE
while (~scanf("%d",&n))
{
scanf("%d",&m);
for (int i=0;i<m;i++)
scanf("%d",X+i);
sort(X,X+m);
bfs();
}
return 0;
}
POJ Multiple (BFS，同余定理),布布扣,bubuko.com
原文地址：http://blog.csdn.net/u012965890/article/details/38376061
