Multiplication table

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 438 Accepted Submission(s): 207

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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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。

这个乘法表不正确，是因为有熊孩子把数字打乱了，样例中，0被换成了1, 1被换成了3, 2被换成了2, 3被换成了0，第一行的0*0等于23其实是3*3=21（四进制）。
求各个数的映射。

题解：探寻乘法表的规律。
数字太大，不能暴搜，而且乘法表是有规律的，我们可以根据数字出现的种类数来判断数字。
样例中有一行全是1，我们可以想到1其实就是0,。
最后一行的高位全是1，我们根据10进制的乘法表来观察，什么时候高位会不变呢？只有0*x和1*x高位是不变的，2*x的时候高位就有0或者1了。所以这个高位全是1的行代表的数字是1。
然后探寻普遍的规律：高位出现了多少种数字，这一位就是多少。（0、1高位都只出现1位数字，需特殊判断两个低位是否相等，相等说明是0，不等是1）

（这题有点碉，我是没想到这么好的判断方法，当时暴搜了起来，果然不得力）

代码（一秒多过，加了读入输出优化的话可以到三百多毫秒）：

 //#pragma comment(linker, "/STACK:102400000,102400000")

 #include<cstdio>

 #include<cmath>

 #include<iostream>

 #include<cstring>

 #include<algorithm>

 #include<cmath>

 #include<map>

 #include<set>

 #include<stack>

 #include<queue>

 using namespace std;

 #define ll long long

 #define usint unsigned int

 #define mz(array) memset(array, 0, sizeof(array))

 #define minf(array) memset(array, 0x3f, sizeof(array))

 #define REP(i,n) for(i=0;i<(n);i++)

 #define FOR(i,x,n) for(i=(x);i<=(n);i++)

 #define RD(x) scanf("%d",&x)

 #define RD2(x,y) scanf("%d%d",&x,&y)

 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)

 #define WN(x) printf("%d\n",x);

 #define RE  freopen("D.in","r",stdin)

 #define WE  freopen("1biao.out","w",stdout)

 #define mp make_pair

 int a[][][];

 int c[];

 int d[];

 int p;

 void farm() {

     int i,j;

     memset(c,-,sizeof(c));

     for(i=; i<p; i++) {

         memset(d,,sizeof(d));

         int cnt=;

         for(j=; j<p; j++)

             if(d[ a[i][j][] ]==) {

                 d[ a[i][j][] ]=;

                 cnt++;

             }

         if(cnt==) cnt= a[i][][]==a[i][][]?:;

         //b[i]=cnt;

         c[cnt]=i;

     }

     for(i=; i<p; i++) {

         printf(" %d",c[i]);

     }

     puts("");

 }

 int main() {

     int cas=, ans;

     int i,j;

     while(scanf("%d",&p)!=EOF) {

         if(p==)break;

         for(i=; i<p; i++) {

             for(j=; j<p; j++) {

                 //scanf("%d%d",&a[i][j][1],&a[i][j][0]);

                 scanf("%d%d",&a[i][j][],&a[i][j][]);

             }

         }

         printf("Case #%d:",cas++);

         farm();

     }

     return ;

 }

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hdu4951 Multiplication table （乘法表的奥秘）

Multiplication table

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