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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.
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Sample Input |
Sample Output |
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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 ",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 ",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(" "); 133 134 } 135 return 0; 136 }