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2012Chhengdu K - Yet Another Multiple Problem
K - Yet Another Multiple Problem
Time Limit:20000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uAppoint description:
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?
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 ≤ 10 4). The second line contains m decimal digits separated by spaces.
Input is terminated by EOF.
For each test case, there are two lines. The first line contains two integers n and m (1 ≤ n ≤ 10 4). 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
题意：以样例为例，2345的最小倍数，不包含给出的三个数7 8 9
思路：bfs，以0-9中能够使用的数字bfs，一位一位的加在后面，当第一个出现数字x%n==0时，则x为解。
这里需要的知识点是，(x*10+i)%n == (x%n)*10+i,所以只需要存(x%n)所有可能，根据抽屉原理，节点数不超过n,这样就可以很快搜到了。
存结果的话，因为结果有可能很长很长，可以在结构体里面加一个字符串，从前面的点的字符串更新过来，也就是在最后加一个‘i‘，或者开数组存这个数的结尾num[i],然后和前面更新过来的节点pre[i].
注意：只有0为可行数字的情况的特殊处理
错在一开始vis数组在第一个数时没置1，只有0为可行数字的情况的特殊处理处理错，还有新的数没%n就放越界。
数组bfs：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define M(a,b) memset(a,b,sizeof(a)) 6 #define INF 0x3f3f3f3f 7 8 using namespace std; 9 10 int n,m;11 12 int pre[1000005],num[1000005];13 int can[15];14 int que[1000005];15 int vis[1000005];16 int res[1000005];17 18 int bfs()19 {20 int head = -1;21 int tail = 0;22 M(vis,0);23 que[0] = 0;24 for(int i = 1;i<10;i++)25 {26 if(!can[i])27 {28 if(i%n==0) {num[i] = i,pre[i] = 0; return i;}29 else num[i] = i,pre[i] = 0, vis[i] = 1, que[tail] = i,tail++;30 }31 }32 while(head<tail)33 {34 head++;35 int tmp = que[head];36 //cout<<tmp<<endl;37 for(int i = 0;i<10;i++)38 {39 if(i==0&&tmp==0) continue;40 //cout<<i<<endl;41 if(!can[i])42 {43 int u = tmp*10+i;44 int t = u%n;45 if(!vis[t])46 {47 if(t==0) {pre[u] = tmp, num[u] = i;return u;}48 else{49 //cout<<t<<‘ ‘<<i<<endl;50 pre[t] = tmp, num[t] = i;51 vis[t] = 1;52 que[tail] = t;53 tail++;54 }55 }56 }57 }58 }59 return -1;60 }61 62 int main()63 {64 int cas = 1;65 while(scanf("%d%d",&n,&m)==2)66 {67 M(pre,0);68 M(num,0);69 M(can,0);70 for(int i = 0;i<m;i++)71 {72 int a;73 scanf("%d",&a);74 can[a] = 1;75 }76 pre[0] = -1;77 int ans = bfs();78 if(m==10) {printf("Case %d: -1\n",cas++); continue;}79 printf("Case %d: ",cas++);80 if(ans == -1) puts("-1");81 else{82 int cnt = 0;83 for(int i = ans;pre[i]!=-1;i = pre[i])84 res[cnt++] = num[i];85 for(int i = cnt-1;i>0;i--)86 printf("%d",res[i]);87 printf("%d\n",res[0]);88 }89 }90 return 0;91 }
queue+struct：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #define M(a,b) memset(a,b,sizeof(a)) 7 #define INF 0x3f3f3f3f 8 9 using namespace std;10 11 int n,m;12 13 struct node{14 int num;15 string c;16 };17 queue<node> que;18 int can[15];19 int vis[1000005];20 int res;21 22 node bfs()23 {24 while(!que.empty()) que.pop();25 M(vis,0);26 for(int i = 1;i<10;i++)27 {28 if(!can[i])29 {30 node tp;31 tp.c = "";32 tp.num = 0;33 if(i%n==0) {char ch = i+‘0‘; tp.c += ch; return tp;}34 else {35 tp.num = i%n;36 char ch = i+‘0‘;37 tp.c += ch;38 vis[i] = 1;39 que.push(tp);40 //cout<<tp.c<<endl;41 }42 }43 }44 while(!que.empty())45 {46 node tmp = que.front();47 que.pop();48 //cout<<tmp.c<<endl;49 for(int i = 0;i<10;i++)50 {51 if(!can[i])52 {53 int t = (tmp.num*10+i)%n;54 if(!vis[t])55 {56 if(t==0) {char ch = i+‘0‘; tmp.c+=ch; return tmp;}57 else58 {59 node tp;60 tp.num = t;61 char ch = i+‘0‘;62 tp.c = tmp.c+ch;63 //cout<<tp.num<<endl;64 vis[t] = 1;65 que.push(tp);66 }67 }68 }69 }70 }71 res = -1;72 node none;73 return none;74 }75 76 int main()77 {78 int cas = 1;79 while(scanf("%d%d",&n,&m)==2)80 {81 res = 0;82 M(can,0);83 for(int i = 0;i<m;i++)84 {85 int a;86 scanf("%d",&a);87 can[a] = 1;88 }89 if(m==10) {printf("Case %d: -1\n",cas++); continue;}90 node ans = bfs();91 printf("Case %d: ",cas++);92 if(res == -1) puts("-1");93 else cout<<ans.c<<endl;94 }95 return 0;96 }
queue+pair：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #define M(a,b) memset(a,b,sizeof(a)) 7 #define INF 0x3f3f3f3f 8 9 using namespace std;10 11 int n,m;12 13 int can[15];14 int vis[1000005];15 int res;16 17 queue<pair<string,int> > rec;18 19 string bfs()20 {21 while (!rec.empty()) rec.pop();22 pair<string,int>init;23 init.first="";init.second=0;24 rec.push(init);25 int i;26 while (!rec.empty())27 {28 pair<string,int> curr=rec.front();29 for (i=0;i<10;i++)30 {31 if (curr.first.length()==0&&i==0) continue;32 if (can[i]) continue;33 char ch=‘0‘+i;34 string ss=curr.first+ch;35 int x=(curr.second*10+i)%n;36 if (!vis[x])37 {38 if (x==0) return ss;39 pair<string,int>u;40 u.first=ss;u.second=x;41 rec.push(u);42 vis[x]=1;43 }44 }45 rec.pop();46 }47 return "-1";48 }49 50 int main()51 {52 int cas = 1;53 while(scanf("%d%d",&n,&m)==2)54 {55 M(can,0);56 M(vis,0);57 for(int i = 0;i<m;i++)58 {59 int a;60 scanf("%d",&a);61 can[a] = 1;62 }63 string ans = bfs();64 printf("Case %d: ",cas++);65 cout<<ans<<endl;66 }67 return 0;68 }
2012Chhengdu K - Yet Another Multiple Problem
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