HDU 4951 Multiplication table(找规律)

题目大意：给出一个p进制数的乘法表，第i行第j组（一组有两个数）表示i*j结果的高位和低位，例如样例中：
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
第二行的“2 3”表示0*0的结果为23，后面的“1 1”表示0*1的结果为11，依此类推。

写出九九乘法表之后你会发现当i*j中，所有高位的数字为0-i-1。然后就可以哈希统计出来高位有多少个，那这一行就是几了啊。

Multiplication table

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 736    Accepted Submission(s): 335

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

42 3 1 1 3 2 1 01 1 1 1 1 1 1 13 2 1 1 3 1 1 21 0 1 1 1 2 1 30

 

Sample Output

Case #1: 1 3 2 0

 

Author

xudyh

 

Source

2014 Multi-University Training Contest 8

 

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-10///#define M 1000100///#define LL __int64#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 510;int num[maxn];int mp[maxn][maxn][2];int vis[maxn];int main(){    int Case = 1;    int p;    while(~scanf("%d",&p) && p)    {        for(int i = 0; i < p; i++)            for(int j = 0; j < p; j++) scanf("%d %d",&mp[i][j][1], &mp[i][j][0]);        for(int i = 0; i < p; i++)        {            int cnt = 0;            memset(vis, 0, sizeof(vis));            for(int j = 0; j < p; j++)            {                if(!vis[mp[i][j][1]])                {                    vis[mp[i][j][1]] = 1;                    cnt++;                }            }            if(cnt == 1)                if(mp[i][0][0] == mp[i][1][0]) cnt = 0;            num[cnt] = i;        }        printf("Case #%d: ", Case++);        for(int i = 0; i < p-1; i++) printf("%d ",num[i]);        printf("%d\n",num[p-1]);    }    return 0;}