題目描述 Description 小明玩一個數字游戲,取個n行n列數字矩陣(其中n為不超過100的奇數),數字的填補方法為:在矩陣中心從1開始以逆時針方向繞行,逐圈擴大,直到n行n列填滿數字,請輸出該n行n列正方形矩陣以及其的對角線數字之和. 輸入描述 Input Description n(即n行n列) 輸出描述 Output Description n+1行,n行為組成的矩陣,最後一行為對角線數字之和 樣例輸入 Sample Input 3 樣例輸出 Sample Output 5 4 3 6 1 2 7 8 9 25
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<bitset> #include<iomanip> using namespace std; int map[ 105 ][ 105 ] ; int main() { int n ; cin >> n ; for(long i = 0; i != n + 1 ; i++ ) { map[ 0 ][ i ] = -1 ; map[ n + 1 ][ i ] = -1 ; map[ i ][ 0 ] = -1 ; map[ i ][ n + 1 ] = -1; } long x = n , y = x ,d = 0; long dir[ 4 ][ 2 ]={ { 0 , -1 },{ -1 , 0 },{ 0 , 1 },{ 1 , 0 }}; for(long i = n * n ; i >= 1 ; i-- ) { map[ x ][ y ] = i; x += dir[ d ][ 0 ]; y += dir[ d ][ 1 ]; if( map[ x ][ y ] != 0 ) { x -= dir[ d ][ 0 ]; y -= dir[ d ][ 1 ]; d = ( d + 1 ) % 4 ; x += dir[ d ][ 0 ]; y += dir[ d ][ 1 ]; } } for( int i = 1 ; i <= n ; ++i ) { for( int j = 1 ; j <= n ; ++j ) cout << map[ i ][ j ] << " " ; cout << endl ; } int ans = 0 ; for( int i = 1 ; i <= n ; ++i ) { ans += map[ i ][ i ] + map[ n - i + 1][ i ] ; } cout << ans - 1 << endl ; return 0 ; } #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<bitset> #include<iomanip> using namespace std; int map[ 105 ][ 105 ] ; int main() { int n ; cin >> n ; for(long i = 0; i != n + 1 ; i++ ) { map[ 0 ][ i ] = -1 ; map[ n + 1 ][ i ] = -1 ; map[ i ][ 0 ] = -1 ; map[ i ][ n + 1 ] = -1; } long x = n , y = x ,d = 0; long dir[ 4 ][ 2 ]={ { 0 , -1 },{ -1 , 0 },{ 0 , 1 },{ 1 , 0 }}; for(long i = n * n ; i >= 1 ; i-- ) { map[ x ][ y ] = i; x += dir[ d ][ 0 ]; y += dir[ d ][ 1 ]; if( map[ x ][ y ] != 0 ) { x -= dir[ d ][ 0 ]; y -= dir[ d ][ 1 ]; d = ( d + 1 ) % 4 ; x += dir[ d ][ 0 ]; y += dir[ d ][ 1 ]; } } for( int i = 1 ; i <= n ; ++i ) { for( int j = 1 ; j <= n ; ++j ) cout << map[ i ][ j ] << " " ; cout << endl ; } int ans = 0 ; for( int i = 1 ; i <= n ; ++i ) { ans += map[ i ][ i ] + map[ n - i + 1][ i ] ; } cout << ans - 1 << endl ; return 0 ; }