ZOJ 1136 Multiple(BFS + 数论 同余剪枝 搜索数字的倍数 )

ZOJ Problem Set - 1136
Multiple
Time Limit: 10 Seconds Memory Limit: 32768 KB

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).

The input file 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.

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:

Input

22
3
7
0
1

2
1
1

Output

110
0
Source: Southeastern Europe 2000
Submit Status
主要是剪枝：数论的剪枝，直接用例子解释，比如36%24==12==60%24，那么36*****%24==60******%24
比如 368%24==608%24，所以在搜索的时候，如果已经搜索到36%24==12，那么遇到60时可以直接跳过

#include<cstdio>
#include<cstring>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#include<algorithm>
#include<queue>
using namespace std;
#define cl(a,b) memset(a,b,sizeof(a))
#define LL long long
const int maxn=100002;
const int inf=9999999;
int n,m;
int a[maxn];
int mod[maxn];
int getMode(string s){
    int len=s.size();
    int ans=0;
    for(int i=0;i<len;i++){
        ans=(ans*10+s[i]-‘0‘)%n;
    }
    return ans;
}
string bfs(){
    cl(mod,0);
    queue<string> q;
    for(int i=0;i<m;i++){//初始化队列的起点
        q.push(string(1,a[i]+‘0‘));//string(number,char)构造函数
    }
    while(!q.empty()){
        string s=q.front();q.pop();
        if(s!="0"&&getMode(s)==0){
            return s;
        }
        for(int i=0;i<m;i++){
            string tmp=s;
            if(tmp=="0"&&a[i]==0)continue;//特判，防止出现00000...的情况
            tmp+=string(1,a[i]+‘0‘);
            int d=getMode(tmp);
            if(mod[d]++)continue;//出现同余，直接跳过
            q.push(tmp);
        }
    }
    return "0";
}

int main(){

    while(~scanf("%d",&n)){
        scanf("%d",&m);
        for(int i=0;i<m;i++){
            scanf("%d",&a[i]);
        }
        if(n==0){
            printf("0\n");continue;
        }
        sort(a,a+m);
        printf("%s\n",bfs().c_str());
    }
    return 0;
}