HDU 4951 Multiplication table 数论

最新推荐文章于 2020-07-09 13:40:13 发布

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本文深入探讨了如何通过已知的混乱乘法表来还原原始基于p进制的乘法表。通过一系列数学操作和逻辑推理，作者详细解释了解决此类问题的方法和步骤，旨在帮助读者理解并解决类似复杂数学问题。

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

可以抽象为10进制找到规律，输入输出数据太多，需要用到输入挂，否则会T

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <set>
#include <vector>
#include <algorithm>
#define maxn 510
using namespace std;

int mark[maxn];
void read(int &a) {
    int t;
    while (t = getchar(), isspace(t));
    a = t - '0';
    while (t = getchar(), !isspace(t)) a = a * 10 + t - '0';
}
int mk[maxn],cnt[maxn],get_zero[maxn];
int main()
{
	//freopen("1007.in","r",stdin);
	//freopen("data.out","w",stdout);
	int n;
	int cas=0;
	while (~scanf("%d",&n))
	{	
		for(int i=0;i<n;i++)
		{
			cnt[i]=0;
		}
		if(n==0) break;
		int tp,tp2;
		
		printf("Case #%d:",++cas);
		for(int i=0;i<n;i++)
		{
			int bl=0;
			memset(mk,-1,sizeof(mk));
			for(int j=0;j<n;j++)
			{
				read(tp);read(tp2);
				//scanf("%d %d",&tp,&tp2);
				if(mk[tp]!=tp){
					cnt[i]++;
					mk[tp]=tp;
				}
				if(mk[tp2]!=tp2)
				{
					bl=1;
				}
			}
			if(!bl&&cnt[i]==1) {
				cnt[i]=0;
			}
			
		}
		for(int i=0;i<n;i++)
		{
	//		printf("%d\n",cnt[i]);
			mark[cnt[i]]=i;
		}
		for(int i=0;i<n;i++)
		{
			printf(" %d",mark[i]);
		}
		puts("");
		
	}
	return 0;
}