HDU 4474 （Yet Another Multiple Problem）同余模定理·

Yet Another Multiple Problem

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6311    Accepted Submission(s): 1465

Problem Description

There are tons of problems about integer multiples. Despite the fact that the topic is not original, the content is highly challenging. That’s why we call it “Yet Another Multiple Problem”.
In this problem, you’re asked to solve the following question: Given a positive integer n and m decimal digits, what is the minimal positive multiple of n whose decimal notation does not contain any of the given digits?

 

Input

There are several test cases.
For each test case, there are two lines. The first line contains two integers n and m (1 ≤ n ≤ 10⁴). The second line contains m decimal digits separated by spaces.
Input is terminated by EOF.

 

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) while Y is the minimal multiple satisfying the above-mentioned conditions or “-1” (without quotation marks) in case there does not exist such a multiple.

 

Sample Input

2345 37 8 9100 10

 

Sample Output

Case 1: 2345Case 2: -1

 

Source

2012 Asia Chengdu Regional Contest

 

【同余模定理】

对于大数求余，超出 （long long int 后） 直接用%  是不行的

但是有这样的定理

（a+b）%c = （a%c + b%c）%c

（a*b）%c = （a%c * b%c ）%c

所以对于大数，我们可以这样：

例如  1234 % 9

1 % 9 = 1

1*10 +2 = 12 % 9= 3

3 * 10 + 3 =33 % 9= 6

6 * 10 + 4 =64 % 9= 1

所以1234 % 9 = 1；  

就有   （x *n + d） % n  求法

【此题题意】

 给n ，m   输入m个数

问 n的最小倍数 不含 这m 个数的  是多少

【思路】

 bfs 搜索；   利于同余模定理  优化

【代码实现】

//#include <bits/stdc++.h>#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <math.h>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n)   scanf("%I64d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)#define Pr(n)     printf("%d\n",n)#define lson rt << 1, l, mid#define rson rt << 1|1, mid + 1, rusing namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int MAXN=10005;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};int n,m;int mods[MAXN];int pre[MAXN];int vis[10];void cprint(int x){    if(pre[x]!=-1)    {        cprint(pre[x]);        //printf("%d",mods[x]);    }    printf("%d",mods[x]);}void bfs(){    queue<int>Q;    for(int i=1;i<10;i++)    {        if(!vis[i]&&mods[i%n]==-1)        {            mods[i%n]=i;            Q.push(i%n);        }    }    while(!Q.empty())    {        int u=Q.front();Q.pop();        if(u==0)// 结束        {            cprint(u);            printf("\n");            return ;        }        for(int i=0;i<10;i++)        {            if(vis[i])                continue;            int x=(u*10+i)%n;            if(mods[x]==-1)//没有被访问过            {                mods[x]=i;                pre[x]=u;                Q.push(x);            }        }    }    printf("-1\n");}int main(){    int x;    int cont=0;    while(~scanf("%d %d",&n,&m))    {        mem(mods,-1);        mem(pre,-1);        mem(vis,0);        while(m--)        {            scanf("%d",&x);            vis[x]=1;        }        printf("Case %d: ",++cont);        bfs();    }    return 0;}

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