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HDOJ 4915 Multiplication table

2024-07-22 00:35:15   212人阅读


可以特判出0和1,再统计每行出现了多少种十位上的数就可以了....

没有考虑2的情况似乎也过了......


Multiplication table

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 800    Accepted Submission(s): 360


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
 

Author
xudyh
 

Source
2014 Multi-University Training Contest 8
 


#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>

using namespace std;

int nextInt()
{
	char ch;
	int ret=0;
	while(ch=getchar())
	{
		if(ch>='0'&&ch<='9')
		{
			ret=ret*10+ch-'0';
		}
		else break;
	}
	return ret;
}

int n,f[510][1010],ans[550],cas=1;
bool v1[510],v2[510];
int s1,s2;

int main()
{
	while(scanf("%d",&n)!=EOF&&n)
	{
		getchar();
		for(int i=0;i<n;i++)
		{
			s1=0; s2=0;
            for(int j=0;j<n;j++)
                v1[j]=v2[j]=true;
			for(int j=0;j<n;j++)
			{
				int num1=nextInt();
				int num2=nextInt();
				if(v1[num1])
                {
                    s1++; v1[num1]=0;
                }
                if(v2[num2])
                {
                    s2++; v2[num2]=0;
                }
			}
			int diff1=s1;
			int diff2=s2;
			if(diff1==1)
			{
				if(diff2==1) ans[0]=i;
				else ans[1]=i;
			}
			else
			{
				int d=diff1;
				ans[d]=i;
			}
		}
		printf("Case #%d:",cas++);
		for(int i=0;i<n;i++)
		{
			printf(" %d",ans[i]);
		}
		putchar(10);
	}
	return 0;
}




HDOJ 4915 Multiplication table


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