【BFS】【余數剪枝】Multiple
阿新 • • 發佈:2017-06-07
wid def ever ems 如何 decimal following exists ota [poj1465]Multiple
On the first line - the number N
On the second line - the number M
On the following M lines - the digits X1,X2..XM.
An example of input and output:
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
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
題目大意:給你一個在0至4999之間的數N,在給你M個數,輸出這些數能組成的最小的N的倍數,沒有輸出0
試題分析:本體盲目搜是不可取的,因為我們根本就不知道要搜到哪裏(對於DFS),每個數的數量可以取無限多個……
那麽對於BFS來說數量還是比較多,但BFS擅長解決一些沒有邊界的問題嘛,所以用BFS解決此題
那麽如何剪枝呢?
對於一個數(AX+Y)%X與(BX+Y)%X的余數是一樣的,至於取誰,小的那個(也就是先搜到的那個)是我們要的,大的可以直接剪掉,所以只需要開一個數組Hash一下就好了……
註意特判N=0的情況!!!
代碼
#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當前數字 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】【余數剪枝】Multiple