POJ-1465（同余+高精度）

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其实很简单的思想，外加贴一个模版

【题目描述】给一个数N 然后给M个一位数问你是否有N的倍数完全由这些一位数组成

先说算法用BFS不停的扩展就是X10这样的扩展然后如果对N取余的余数没有出现过就把这个扩展得数的余数添加到队列里如果余数是0的话就可以输出了
当然扩展的时候要考虑到0，这些都不是最关键的最关键的是这个数可能非常大 long long 不够而高精的话比较麻烦。用链表。
还有一点这个队列最多只有5000就够了。
另外不需要证明所有的余数都取到了没有必要

/******************************************************************************************************

 ** Copyright (C) 2011.07.01-2013.07.01 ** Author: famousDT <13730828587@163.com> ** Edit date: 2012-03-11******************************************************************************************************/#include <stdio.h>#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10#include <vector>#include <queue>#include <deque>//two direction queue#include <map>#include <time.h>#include <set>#include <list>#include <stack> #include <string>#include <iostream>#include <fstream>#include <assert.h>#include <bitset>#include <iterator>//C++Primer#include <string.h>//memcpy(to,from,count#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll#include <numeric>#include <utility>//pair#include <functional>#include <algorithm>using namespace std;typedef long long int ll;#define MY_E exp(1.0)#define MY_PI acos(-1.0)#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))#define MY_INT_MAX 0x7fffffff#define CLEAR(a) memset(a, 0, sizeof(a))/*==========================================================*\| \*==========================================================*/ #define N 5000 + 5struct node{int res; //余数int digit; //位数node *pre;} q[N];bool hash[N];void output(node x){if (x.pre) {output(*x.pre);printf("%d", x.digit);}}void bfs(int n, int m, int d[]){int i, j;CLEAR(hash);int front = 0, rear = 1;q[0].pre = NULL;q[0].digit = q[0].res = 0;node * first = &q[0];while (front < rear) {node now, curr = q[front];now.pre = &q[front];for (i = 0; i < m; ++i) {int res = (curr.res * 10 + d[i]) % n;if (hash[res] == 0 && (now.pre != first || d[i] > 0)) {hash[res] = true;now.res = res;now.digit = d[i];q[rear++] = now;if (now.res == 0) {output(now);printf("\n");return;}}}++front;}printf("0\n");}int main(){int m, n;int d[15];int i;while (scanf("%d%d", &n, &m) == 2) {for (i = 0; i < m; ++i)scanf("%d", &d[i]);if (!n) {printf("0\n");continue;}sort(d, d + m);bfs(n, m, d);}return 0;}