[poj1465]Multiple
| Time Limit: 1000MS | Memory Limit: 32768K | |
| Total Submissions: 7731 | Accepted: 1723 |
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
题目大意：给你一个在0至4999之间的数N，在给你M个数，输出这些数能组成的最小的N的倍数，没有输出0
试题分析：本体盲目搜是不可取的，因为我们根本就不知道要搜到哪里(对于DFS)，每个数的数量可以取无限多个……
那么对于BFS来说数量还是比较多，但BFS擅长解决一些没有边界的问题嘛，所以用BFS解决此题
那么如何剪枝呢？
对于一个数(AX+Y)%X与(BX+Y)%X的余数是一样的，至于取谁，小的那个(也就是先搜到的那个)是我们要的，大的可以直接剪掉，所以只需要开一个数组Hash一下就好了……
注意特判N=0的情况！！！
代码
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
//#include<cmath>
using namespace std;
const int INF = 9999999;
#define LL long long
inline int read(){
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
return x*f;
}
int N,M;
int a[5101];
int l=1,r=1;
bool flag[5101];
bool fla=false;
struct data{
int ga,last,ans;
}Que[5101];
void print(data k){
if(k.ga!=-1){
print(Que[k.ga]);
printf("%d",k.ans);
}
}
void BFS(){//a当前数字
Que[1].last=0;
Que[1].ans=0;
Que[1].ga=-1;
int l1,p;
while(l<=r){
l1=Que[l].last;
for(int i=1;i<=M;i++){
p=(l1*10+a[i])%N;
if(!flag[p]&&(Que[l].ga!=-1||a[i]>0)){
flag[p]=true;
Que[++r].ans=a[i];
Que[r].ga=l;
Que[r].last=p;
if(p==0){
data s=Que[r];
print(s);
printf("\n");
fla=true;
return ;
}
}
}
l++;
}
}
int main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
while(scanf("%d",&N)!=EOF){
M=read();
for(int i=1;i<=M;i++) a[i]=read();
sort(a+1,a+M+1);
if(N==0){
puts("0");
continue;
}
memset(flag,false,sizeof(flag));
l=r=1;
fla=false;
BFS();
if(!fla){
puts("0");
}
}
return 0;
}