UVA 11754 Code Feat （枚举，中国剩余定理）

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C

Code Feat

  

 

The government hackers at CTU (Counter-Terrorist Unit) have learned some things about the code, but they still haven't quite solved it.They know it's a single, strictly positive, integer.  They also know several clues of the form "when divided by X, the remainder is one of {Y1, Y2, Y3, ...,Yk}".There are multiple solutions to these clues, but the code is likely to be one of the smallest ones.  So they'd like you to print out the first few solutions, in increasing order.

The world is counting on you!

Input

Input consists of several test cases.  Each test case starts with a line containing C, the number of clues (1 <= C <= 9), and S, the number of desired solutions (1 <= S <= 10).  The next C lines each start with two integers X (2 <= X) and k (1 <= k <= 100), followed by the k distinct integers Y1, Y2, ...,Yk (0 <= Y1, Y2, ..., Yk < X).

You may assume that the Xs in each test case are pairwise relatively prime (ie, they have no common factor except 1).  Also, the product of the Xs will fit into a 32-bit integer.

The last test case is followed by a line containing two zeros.

Output

For each test case, output S lines containing the S smallest positive solutions to the clues, in increasing order.

Print a blank line after the output for each test case.

Sample Input                              

Sample Output

 

3 2

2 1 1

5 2 0 3

3 2 1 2

0 0

5

13

  

 

Problem Setter: Derek Kisman, Special Thanks: SameeZahur

当所有k的乘积较小时，直接枚举出所有的情况，然后用中国剩余定理（CRT）即可求解；

当所有k的乘积较大时，直接枚举所有值，判断是否符合即可。注意k/x越小越好，这样t*x+yi足够大，很快就能找到。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <set>
  4 #include <cstring>
  5 #include <vector>
  6 #include <algorithm>
  7 using namespace std;
  8 long long X[11],K[11],Y[11][110];
  9 vector<long long >vec;
 10 set<int>val[11];
 11 long long c,s;
 12 long long mod;
 13 int minn=0;
 14 typedef long long ll;
 15 void ext_gcd(ll a,ll b,ll &d,ll &x,ll &y)
 16 {
 17     if(!b){d=a;x=1;y=0;}
 18     else
 19     {
 20         ext_gcd(b,a%b,d,y,x);
 21         y-=x*(a/b);
 22     }
 23 }
 24 ll a[11];
 25 ll CRT()
 26 {
 27     ll d,x,y,ret=0;
 28     ll temp;
 29     for(int i=0;i<c;i++)
 30     {
 31         temp=mod/X[i];
 32         ext_gcd(X[i],temp,d,x,y);
 33         ret=(ret+y*temp*a[i])%mod;
 34     }
 35     return (ret+mod)%mod;
 36 }
 37 void dfs(int d)
 38 {
 39     if(d==c)vec.push_back(CRT());
 40     else
 41     {
 42         for(int i=0;i<K[d];i++)
 43         {
 44             a[d]=Y[d][i];
 45             dfs(d+1);
 46         }
 47     }
 48 }
 49 void solve1()
 50 {
 51     vec.clear();
 52     dfs(0);
 53     sort(vec.begin(),vec.end());
 54     int size=vec.size();
 55     int num=0;
 56     for(int i=0;;i++)
 57     {
 58         for(int j=0;j<size;j++)
 59         {
 60             ll ans=mod*i+vec[j];
 61             if(ans>0)
 62             {
 63                 printf("%lld\n",ans);
 64                 num++;
 65                 if(num==s)return ;
 66             }
 67         }
 68     }
 69 }
 70 void solve2()
 71 {
 72     for(int i=0;i<c;i++)
 73     {
 74         if(i!=minn)
 75         {
 76             val[i].clear();
 77             for(int j=0;j<K[i];j++)
 78             {
 79                 val[i].insert(Y[i][j]);
 80             }
 81         }
 82     }
 83     ll ans=0;
 84     bool ok=1;
 85     int num=0;
 86     for(int i=0;;i++)
 87     {
 88         for(int j=0;j<K[minn];j++)
 89         {
 90             ans=X[minn]*i+Y[minn][j];
 91             if(ans<=0)continue;
 92             ok =1;
 93             for(int k=0;k<c;k++)
 94             {
 95                 if(k!=minn&&!val[k].count(ans%X[k]))
 96                 {
 97                     ok=0;
 98                     break;
 99                 }
100             }
101             if(ok)
102             {
103                 printf("%lld\n",ans);
104                 num++;
105                 if(num==s)return;
106             }
107         }
108     }
109 }
110 int main()
111 {
112     while(scanf("%lld%lld",&c,&s)==2&&(c||s))
113     {
114         if(c==0&&s==0)break;
115         mod=1;
116         minn=0;
117         long long k=1;
118         for(int i=0;i<c;i++)
119         {
120             scanf("%lld%lld",&X[i],&K[i]);
121             mod*=X[i];
122             k*=K[i];
123             for(int j=0;j<K[i];j++)
124             {
125                 scanf("%lld",&Y[i][j]);
126             }
127             sort(Y[i],Y[i]+K[i]);
128             if(K[i]*X[minn]>K[minn]*X[i])minn=i;
129         }
130         if(k>10000)solve2();
131         else solve1();
132         printf("\n");
133 
134     }
135     return 0;
136 }
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posted on 2014-11-27 22:24  xyiyy  阅读(424)  评论(0)  编辑  收藏  举报

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