Multiplication table

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 435    Accepted Submission(s): 204

Problem Description

Teacher Mai has a multiplication table in base p.
   For example, the following is a multiplication table in base 4:*  0  1  2  3 0 00 00 00 00 1 00 01 02 03 2 00 02 10 12 3 00 03 12 21   But a naughty kid maps numbers 0..p-1 into another permutation and shuffle the multiplication table.
   For example Teacher Mai only can see:1*1=11 1*3=11 1*2=11 1*0=11 3*1=11 3*3=13 3*2=12 3*0=10 2*1=11 2*3=12 2*2=31 2*0=32 0*1=11 0*3=10 0*2=32 0*0=23   Teacher Mai wants you to recover the multiplication table. Output the permutation number 0..p-1 mapped into.
   It‘s guaranteed the solution is unique.

Input

There are multiple test cases, terminated by a line "0".
   For each test case, the first line contains one integer p(2<=p<=500).
   In following p lines, each line contains 2*p integers.The (2*j+1)-th number x and (2*j+2)-th number y in the i-th line indicates equation i*j=xy in the shuffled multiplication table.
Warning: Large IO!

Output

For each case, output one line.
   First output "Case #k:", where k is the case number counting from 1. The following are p integers, indicating the permutation number 0..p-1 mapped into.

Sample Input

4
2 3 1 1 3 2 1 0
1 1 1 1 1 1 1 1
3 2 1 1 3 1 1 2
1 0 1 1 1 2 1 3
0

Sample Output

Case #1: 1 3 2 0

Source

2014 Multi-University Training Contest 8

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分析：大水题，，，要敢于分析。。。

详见题解：http://blog.sina.com.cn/u/1809706204

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<cstdio>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<map>
 9
10 #define N 1005
11 #define M 15
12 #define mod 1000000007
13 #define mod2 100000000
14 #define ll long long
15 #define maxi(a,b) (a)>(b)? (a) : (b)
16 #define mini(a,b) (a)<(b)? (a) : (b)
17
18 using namespace std;
19
20 int cnt;
21 int p;
22 int i,j;
23 int a[N][N];
24 int c[N][N];
25 int cc[N];
26 int ans[N];
27
28 int main()
29 {
30     //ini();
31     //freopen("data.in","r",stdin);
32     //scanf("%d",&T);
33     //for(int cnt=1;cnt<=T;cnt++)
34     //while(T--)
35     cnt=1;
36     while(scanf("%d",&p)!=EOF)
37     {
38         if(p==0) break;
39         printf("Case #%d:",cnt);cnt++;
40         memset(c,0,sizeof(c));
41         memset(cc,0,sizeof(cc));
42        for(i=0;i<p;i++)
43        {
44            for(j=1;j<=p;j++){
45                 scanf("%d",&a[i][2*j-1]);
46                 cc[ a[i][2*j-1] ]++;
47                 scanf("%d",&a[i][2*j]);
48                 cc[ a[i][2*j] ]++;
49            }
50        }
51
52        int ma=cc[0];
53        int index=0;
54        for(i=0;i<p;i++){
55             if(cc[i]>ma){
56                 ma=cc[i];index=i;
57             }
58        }
59        ans[0]=index;
60
61        memset(cc,0,sizeof(cc));
62
63        for(i=0;i<p;i++)
64        {
65            for(j=1;j<=p;j++){
66                 if(c[i][ a[i][2*j-1] ]==0){
67                     c[i][ a[i][2*j-1] ]=1;
68                     cc[i]++;
69                 }
70            }
71            if(cc[i]==1){
72                 if(ans[0]!=i) ans[1]=i;
73            }
74            else{
75                 ans[ cc[i] ]=i;
76            }
77        }
78
79        for(i=0;i<p;i++) printf(" %d",ans[i]);
80        printf("\n");
81     }
82
83
84
85     return 0;
86 }

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