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;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。