1154:0 starting point learning algorithm 61--matrix transpose

1154:0 starting point learning algorithm 61--matrix transpose time limit:1 Sec Memory limit:64 MB 64bit IO Format:%lld
submitted:1324 accepted:698
[Submit] [Status] [Web Board] Description

You are now asked to transpose a matrix column and output, noting that the number of rows and columns may be different.

Input

Multiple sets of test data, each set of test data first input n and M, indicating the number of rows and columns of the matrix (1 < n,m <= 10)
Then a matrix of n rows m columns

Output

For each set of test data output the matrix after transpose

Sample Input

3 21 23) 45 6

Sample Output

1 3 52) 4 6

Source

0 Starting point Learning algorithm

1#include <stdio.h>2 intMain () {3     intn,m,a[Ten][Ten],b[Ten][Ten];4      while(SCANF ("%d%d", &n,&m)! =EOF) {5         6          for(intI=0; i<n;i++){7              for(intj=0; j<m;j++){8scanf"%d",&a[i][j]);9             }Ten         } One          A          for(intI=0; i<m;i++){ -              for(intj=0; j<n;j++){ -b[i][j]=A[j][i]; the             } -         } -          -          for(intI=0; i<m;i++){ +              for(intj=0; j<n-1; j + +){ -printf"%d", B[i][j]); +             } Aprintf"%d\n", b[i][n-1]); at         } -     } -     return 0; -}

1154:0 starting point learning algorithm 61--matrix transpose