首页 > 代码库 > 【BFS】【余数剪枝】Multiple
【BFS】【余数剪枝】Multiple
[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.
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:
An example of input and output:
Sample Input
223701211
Sample Output
1100
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 longinline 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;}
【BFS】【余数剪枝】Multiple
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。