1 0 1 0
0 0 0 0
1 1 1 1
0 1 0 1
The sums of the rows are 2, 0, 4 and 2. The sums of the columns are 2, 2, 2 and 2.
Your job is to write a program that reads in a matrix and checks if it has the parity property. If not, your program should check if the parity property can be established by changing only one bit. If this is not possible either, the matrix should be classified as corrupt.
Input
The input will contain one or more test cases. The first line of each test case contains one integer n (n < 100), representing the size of the matrix. On the next n lines, there will be n integers per line. No other integers than 0 and 1 will occur in the matrix. Input will be terminated by a value of 0 for n.
Output
For each matrix in the input file, print one line. If the matrix already has the parity property, print "OK". If the parity property can be established by changing one bit, print "Change bit (i,j)" where i is the row and j the column of the bit to be changed. Otherwise, print "Corrupt".
Sample Input
4
1 0 1 0
0 0 0 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 0 1 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 1 1 0
1 1 1 1
0 1 0 1
0
Sample Output
OK
Change bit (2,3)
Corrupt
如果 横着,或者竖着,有两个或以上的奇数和,那么肯定Corrupt
假如横竖各有一个,那么可以更改,则横竖坐标分别是横竖为奇数的那一行/列
所以
代码:
#include <iostream> using namespace std; int main() { int n; while(cin>>n && n) { int matr[n][n]; int sum,x,y; int flag1=0,flag2=0; for(int i=0; i<n; i++) { sum=0; for(int j=0; j<n; j++) { cin>>matr[i][j]; if(matr[i][j]) sum++; } if(sum%2!=0) { flag1++; x=i; } } for(int i=0; i<n; i++) { sum=0; for(int j=0; j<n; j++) { if(matr[j][i]) sum++; } if(sum%2!=0) { flag2++; y=i; } } if(flag1==0&&flag2==0) cout<<"OK"<<endl; else if(flag1==1&&flag2==1) cout<<"Change bit ("<<x+1<<","<<y+1<<")"<<endl; else if(flag1>=2||flag2>=2) cout<<"Corrupt"<<endl; } return 0; }