目前只学了数组之前的基本知识
指针只限于知道  还没学到数组与指针那张
这是作业题  求大佬们用基础的语句解答下
define a m×n array ,output the array with the following type using 1-m*n
输入 3 4
输出
1 2 3 4
10 11 12 5
9 8 7 6

QvQ

这太简单了。
#include<stdio.h>
int main()
{
int a,b;
scanf("%d%d",&a,&b);
printf("1 2 3 4\n10 11 12 5\n9 8 7 6");
return 0;
}

题目不会这么简洁吧？

这标题起得……，很无聊

#include <stdio.h>

void foo( unsigned row, unsigned col )
{
    for( unsigned i=0; i!=row*col; ++i )
    {
        unsigned r = i/col;
        unsigned c = i%col;

        unsigned v;
        if( r-0<=row-1-r && r-0<=c-0 && r-0<=col-1-c ) // →
            v = 2*(row+col-2*r)*r + c-r+1;
        else if( col-1-c<=c-0 && col-1-c<=r-0 && col-1-c<=row-1-r ) // ↓
            v = 2*(row+col-2*(col-1-c))*(col-1-c) + r+3*c-2*col+3;
        else if( row-1-r<=r-0 && row-1-r<=col-1-c && row-1-r<=c-0 ) // ←
            v = 2*(row+col-2*(row-1-r))*(row-1-r) + 2*col-4*row+5*r-c+3;
        else if( c-0<=col-1-c && c-0<=row-1-r && c-0<=r-0 ) // ↑
            v = 2*(row+col-2*c)*c + 2*row+2*col-r-7*c-3;
        printf( "%u%c", v, " \n"[c+1==col] );
    }
}

int main( void )
{
    unsigned m, n;
    scanf( "%u%u", &m, &n );
    foo( m, n );
}

。。。。。你没懂题

相当于你输入矩阵规格，按那种数字方向打印出矩阵，从1开始

标题废
能否说一下思路，我还是有点乱，虽然程序语言看懂了，但我还是不太能想清楚，标量和条件有点绕

跟程序关系不大，就是简单的数学题，根据 行号、列号 求其值。

#include <stdio.h>

void foo( unsigned row, unsigned col, unsigned base, const char* fmt )
{
    for( unsigned i=0; i!=row*col; ++i )
    {
        unsigned T = i/col;
        unsigned B = row-1-T;
        unsigned L = i%col;
        unsigned R = col-1-L;

        unsigned v;
        #define bar(row,col,t,l,base) (2*(row)*(t) + 2*(col)*(t) - 4*(t)*(t) + (l)-(t) + (base))
            if( T<=B && T<=L && T<=R ) // →
                v = bar( row, col, T, L, base );
            else if( R<=T && R<=B && R<=L ) // ↓
                v = bar( col, row-1, R, T-1, base+col );
            else if( B<=T && B<=L && B<=R ) // ←
                v = bar( row-1, col-1, B, R-1, base+col+row-1 );
            else if( L<=T && L<=B && L<=R ) // ↑
                v = bar( col-1, row-2, L, B-1, base+2*col+row-2 );
        #undef bar
        printf( fmt, v );
        putchar( " \n"[R==0] );
    }
}

int main( void )
{
    foo( 5, 4, 1, "%02u" );
    foo( 5, 5, 1, "%02u" );
    foo( 5, 6, 1, "%02u" );
    foo( 7, 11, 1, "%02u" );
    foo( 17, 9, 1, "%03u" );
}