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=1;
    while(scanf("%d",&p)!=EOF)
    {
        if(p==0) break;
        printf("Case #%d:",cnt);cnt++;
        memset(c,0,sizeof(c));
        memset(cc,0,sizeof(cc));
       for(i=0;i<p;i++)
       {
           for(j=1;j<=p;j++){
                scanf("%d",&a[i][2*j-1]);
                cc[ a[i][2*j-1] ]++;
                scanf("%d",&a[i][2*j]);
                cc[ a[i][2*j] ]++;
           }
       }

       int ma=cc[0];
       int index=0;
       for(i=0;i<p;i++){
            if(cc[i]>ma){
                ma=cc[i];index=i;
            }
       }
       ans[0]=index;

       memset(cc,0,sizeof(cc));

       for(i=0;i<p;i++)
       {
           for(j=1;j<=p;j++){
                if(c[i][ a[i][2*j-1] ]==0){
                    c[i][ a[i][2*j-1] ]=1;
                    cc[i]++;
                }
           }
           if(cc[i]==1){
                if(ans[0]!=i) ans[1]=i;
           }
           else{
                ans[ cc[i] ]=i;
           }
       }

       for(i=0;i<p;i++) printf(" %d",ans[i]);
       printf("
");
    }



    return 0;
}