HDU6046-hash

hash

                                                                   Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
                                                                                               Total Submission(s): 502    Accepted Submission(s): 155

Problem Description

Qscqesze is busy at data cleaning.
One day,he generates a large matrix by Jenkins one-at-a-time hash:
inline unsigned sfr(unsigned h, unsigned x) {
  return h >> x;
}
int f(LL i, LL j) {
  LL w = i * 1000000ll + j;
  int h = 0;
  for(int k = 0; k < 5; ++k) {
    h += (int) ((w >> (8 * k)) & 255);
    h += (h << 10);
    h ^= sfr(h, 6);
  }
  h += h << 3;
  h ^= sfr(h, 11);
  h += h << 15;
  return sfr(h, 27) & 1;
}
Obviously,it's a 1e6*1e6 matrix.The data is at row i column j is f(i,j).Note that i and j are both numbered from 1.
Then he gets some matrices sized 1e3*1e3 from the matrix above.But he forgets their original postion.Can you help him to find them out?You just are asked to tell Qscqesze the left-top corner's postion.

 

Input

The first line is the number of test cases T (T<=3). 
Here come with T cases.Each case is consist of 1000 0/1-strings sized 1000.
For convenience,the sample input is 10*10.And the real testcase is 1e3*1e3.

 

Output

For each test case, output a single line "Case #x :y z", where x is the case number, starting from 1. And y z is the answer.

 

Sample Input

  

   1
0000011100
0000110011
0111111100
0011110010
0110101010
1001001001
0100111110
1111001010
0011101110
1100110100
  

 

Sample Output

  

   Case #1 :123456 234567
  

 

Source

2017 Multi-University Training Contest - Team 2

 

题意：给出一个1e3*1e3的小矩阵，让你在一个1e6*1e6的矩形内它，并输出左上端点

解题思路：先在1e3*1e3的小矩阵中处理出所有8*8的识别矩阵。然后遍历1e6*1e6的矩阵，遍历时不需要遍历所有点，只需要行之间隔1000，列之间隔1000即可（因为这样里面至少覆盖了一个识别矩阵），然后处理出这个以这个点位左上端点的8*8的小矩阵，然后判断在1e3*1e3的矩阵中能否找到这个矩阵即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

unordered_map<LL,pair<int,int>>mp;
char ch[1009][1009];

inline unsigned sfr(unsigned h, unsigned x)
{
    return h>>x;
}

int f(LL i, LL j)
{
    LL w=i*1000000ll+j;
    int h=0;
    for(int k=0;k<5;k++)
    {
        h+=(int)((w>>(8*k))&255);
        h+=(h<<10);
        h^=sfr(h,6);
    }
    h+= h<<3;
    h^=sfr(h,11);
    h+=h<<15;
    return sfr(h,27)&1;
}

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=1;i<=1000;i++)scanf("%s",ch[i]+1);
        mp.clear();
        for(int i=1;i<=993;i++)
        {
            for(int j=1;j<=993;j++)
            {
                LL temp=0;
                for(int k=0;k<8;k++)
                    for(int p=0;p<8;p++)
                    {
                        temp<<=1;
                        if(ch[i+k][j+p]=='1') temp|=1;
                    }
                mp[temp]=make_pair(i,j);
            }
        }
        int flag=0;
        for(int i=1;i<=1e6;i+=992)
        {
            for(int j=1;j<=1e6;j+=992)
            {
                LL temp=0;
                for(int k=0;k<8;k++)
                    for(int p=0;p<8;p++)
                    {
                        temp<<=1;
                        if(f(i+k,j+p)) temp|=1;
                    }
                if(mp.find(temp)!=mp.end())
                {
                    pair<int,int> ans=mp[temp];
                    printf("Case #%d :%d %d\n",++cas,i-ans.first+1,j-ans.second+1);
                    flag=1;break;
                }
            }
            if(flag) break;
        }
    }
    return 0;
}