Hooray! Agent Bauer has shot the terrorists, blown upthe bad guy base, saved the hostages, exposed the moles in the government,prevented an environmental catastrophe, and found homes for three orphanedkittens, all in the span of 19 consecutive hours. But now, he only has 5 hours remaining todeal with his final challenge: an activated nuclear bomb protected by asecurity code. Can you help him figureout the code and deactivate it? Eventsoccur in real time.
The governmenthackers at CTU (Counter-Terrorist Unit) have learned some things about thecode, 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 "whendivided by X, the remainder is one of {Y1, Y2, Y3, ...,Yk}".There are multiple solutions to these clues, but the code is likely tobe one of the smallest ones. So they‘dlike you to print out the first few solutions, in increasing order.
The world iscounting on you!
Input
Input consistsof several test cases. Each test casestarts with a line containing C, the number of clues (1 <= C <= 9), andS, the number of desired solutions (1 <= S <= 10). The next C lines each start with two integersX (2 <= X) and k (1 <= k <= 100), followed by the k distinct integersY1, Y2, ..., Yk (0 <= Y1,Y2, ..., Yk < X).
You may assumethat the Xs in each test case are pairwise relativelyprime (ie, they have no common factor except 1). Also, the product of the Xs will fit into a32-bit integer.
The last testcase is followed by a line containing two zeros.
Output
For each testcase, output S lines containing the S smallest positive solutions to the clues,in increasing order.
Print a blankline 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 |
思路：中国剩余定理+暴力；
首先我们肯定会想到dfs暴力枚举，这种情况只有当所有k的乘积很小的情况下才行，这个时候我们采取dfs+中国剩余定理；当k的乘积很大的情况下，
考虑选择一个Ｋ/Ｘ最小的条件，然后枚举所有符合这个条件的数ｎ，即 tX+Yi ，t=0,1,2.... 然后依次判断这个ｎ是否符合其它所有条件，因为这个时候x的增量很大。
1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #include <algorithm>
5 #include <map>
6 #include <vector>
7 #include <set>
8 typedef long long LL;
9 using namespace std;
10 typedef struct node
11 {
12 int x;
13 int y;
14 int bns[105];
15 double b;
16 } ss;
17 set<LL>ask;
18 set<LL>::iterator it;
19 set<int>val[20];
20 ss ans[20];
21 int a[20];
22 void dfs(int x,int y);
23 LL china(int x);
24 pair<LL,LL>ex_gcd(LL n,LL m);
25 void slove__x(int n,int id,int m);
26 LL R;
27 bool cmp(node p,node q)
28 {
29 return p.b < q.b;
30 }
31 int main(void)
32 {
33 int n,m;
34 while(scanf("%d %d",&n,&m)!=EOF)
35 {
36 if(n==0||m==0)break;
37 int i,j;
38 int id = 0;
39 LL um = 1;
40 for(i = 0; i < n; i++)
41 {
42 scanf("%d",&ans[i].x);
43 scanf("%d",&ans[i].y);
44 for(j = 0; j < ans[i].y; j++)
45 {
46 scanf("%d",&ans[i].bns[j]);
47 }
48 sort(ans[i].bns,ans[i].bns+ans[i].y);
49 if(ans[id].x*ans[i].y<ans[i].x*ans[id].y)id=i;
50 um *= ans[i].y;
51 ans[id].b = 1.0*ans[i].y/ans[i].x;
52 }
53 sort(ans,ans+n,cmp);
54 if(um <= 10000)
55 {
56 ask.clear();
57 R = 1;
58 for(i = 0 ; i < n; i++)
59 {
60 R *= ans[i].x;
61 }
62 dfs(0,n);
63 int cn = 0;
64 for(i = 0 ; m!=0; i++)
65 {
66 for(it = ask.begin(); it!=ask.end(); it++)
67 {
68 LL tt = *it + i*R;
69 if(tt > 0)
70 {
71 m--;
72 printf("%lld\n",tt);
73 if(m==0)
74 break;
75 }
76 }
77 }
78 }
79 else
80 {
81 slove__x(n,id,m);
82 }
83 printf("\n");
84 }
85 return 0;
86 }
87 void dfs(int x,int y)
88 {
89 if(x == y)
90 {
91 ask.insert(china(y));
92 return ;
93 }
94 else for(int i = 0; i < ans[x].y; i++)
95 {
96 a[x] = ans[x].bns[i];
97 dfs(x+1,y);
98 }
99 }
100 LL china(int x)
101 {
102 int i,j;
103 LL sum = 0;
104 LL ac = 1;
105 ac = R;
106 for(i = 0; i < x; i++)
107 {
108 LL ak = ac/ans[i].x;
109 pair<LL,LL>cc = ex_gcd(ak,ans[i].x);
110 LL aa = cc.first;
111 aa = (aa%ans[i].x + ans[i].x)%ans[i].x;
112 sum = sum + ((aa*ak%ac)*a[i])%ac;
113 sum %= ac;
114 sum += ac;
115 sum %= ac;
116 }
117 return sum;
118 }
119 pair<LL,LL>ex_gcd(LL n,LL m)
120 {
121 if(m == 0)
122 return make_pair(1,0);
123 else
124 {
125 pair<LL,LL>ac = ex_gcd(m,n%m);
126 return make_pair(ac.second,ac.first-(n/m)*ac.second);
127 }
128 }
129 void slove__x(int n,int id,int m)
130 {
131 int i,j;
132 for(i = 0; i < n; i++)
133 val[i].clear();
134 for(i = 0; i < n; i++)
135 {
136 for(j = 0; j < ans[i].y; j++)
137 {
138 val[i].insert(ans[i].bns[j]);
139 }
140 }
141 for(i = 0; m != 0; i++)
142 {
143 for(j = 0; j < ans[id].y; j++)
144 {
145 LL ak = (LL)i*(LL)ans[id].x + ans[id].bns[j];
146 if(ak > 0)
147 {
148 int flag = 0;
149 for(int c = 0; c < n; c++)
150 {
151 if(c!=id)
152 {
153 if(!val[c].count(ak%ans[c].x))
154 {
155 flag = 1;
156 break;
157 }
158 }
159 }
160 if(!flag)
161 {
162 printf("%lld\n",ak);
163 m--;
164 }
165 if(m==0)break;
166 }
167 }
168 }
169 }