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;
}
HDU 4951 Multiplication table 数论