hdu 4951

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

 

Recommend

hujie   |   We have carefully selected several similar problems for you:  4955 4954 4953 4952 4950 

 

分析：大水题，，，要敢于分析。。。

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

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>

 #define N 1005
 #define M 15
 #define mod 1000000007
 #define mod2 100000000
 #define ll long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int cnt;
 int p;
 int i,j;
 int a[N][N];
 int c[N][N];
 int cc[N];
 int ans[N];

 int main()
 {
     //ini();
     //freopen("data.in","r",stdin);
     //scanf("%d",&T);
     //for(int cnt=1;cnt<=T;cnt++)
     //while(T--)
     cnt=;
     while(scanf("%d",&p)!=EOF)
     {
         if(p==) break;
         printf("Case #%d:",cnt);cnt++;
         memset(c,,sizeof(c));
         memset(cc,,sizeof(cc));
        for(i=;i<p;i++)
        {
            for(j=;j<=p;j++){
                 scanf("%d",&a[i][*j-]);
                 cc[ a[i][*j-] ]++;
                 scanf("%d",&a[i][*j]);
                 cc[ a[i][*j] ]++;
            }
        }

        int ma=cc[];
        int index=;
        for(i=;i<p;i++){
             if(cc[i]>ma){
                 ma=cc[i];index=i;
             }
        }
        ans[]=index;

        memset(cc,,sizeof(cc));

        for(i=;i<p;i++)
        {
            for(j=;j<=p;j++){
                 if(c[i][ a[i][*j-] ]==){
                     c[i][ a[i][*j-] ]=;
                     cc[i]++;
                 }
            }
            if(cc[i]==){
                 if(ans[]!=i) ans[]=i;
            }
            else{
                 ans[ cc[i] ]=i;
            }
        }

        for(i=;i<p;i++) printf(" %d",ans[i]);
        printf("\n");
     }

     return ;
 }