6 solutions
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2
#include <bits/stdc++.h> using namespace std; int n,m; int a[105][105]; int main(){ cin>>n>>m; for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) cin >> a[i][j]; for(int i = 1;i <= m;i++){ for(int j = 1;j <= n;j++) cout << a[j][i] << " "; cout << endl; } return 0; }
秒了
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0
#include <bits/stdc++.h> using namespace std; int main(int argc, char **argv){ int n,m; cin >> n >> m; int a[n + 5][m + 5]; for (int i = 0;i < n;i++){ for (int j = 0;j < m;j++){ cin >> a[i][j]; } } for (int i = 0;i < m;i++){ for (int j = 0;j < n;j++){ printf("%d ",a[j][i]); } cout << endl; } return 0; }
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0
#include <set> #include <ios> #include <list> #include <cmath> #include <ctime> #include <queue> #include <deque> #include <stack> #include <vector> #include <bitset> #include <cctype> #include <cerrno> #include <cwchar> #include <cstdio> #include <fenv.h> #include <iosfwd> #include <string> #include <limits> #include <math.h> #include <cstdlib> #include <iomanip> #include <clocale> #include <complex> #include <cstring> #include <cstring> #include <cwctype> #include <istream> #include <ostream> #include <sstream> #include <fstream> #include <utility> #include <stdio.h> #include <iostream> #include <stdint.h> #include <string.h> #include <tgmath.h> #include <complex.h> #include <algorithm> #include <exception> #include <stdbool.h> #include <stdexcept> #include <streambuf> #include <functional> #include <inttypes.h> #include <bits/stdc++.h> using namespace std; int a[105][105]; int main(){ int n,m; cin >> n >> m; for(int i = 1;i <= n;i++){ for(int j = 1;j <= m;j++){ cin >> a[i][j]; } } for(int i = 1;i <= m;i++){ for(int j = 1;j <= n;j++){ cout << a[j][i] << " "; } cout << endl; } return 0; }`
终于做出来了
- 1
Information
- ID
- 612
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 4
- Tags
- # Submissions
- 118
- Accepted
- 52
- Uploaded By