1 条题解

  • 0
    @ 2025-2-14 20:50:07

    C :

    #include <stdio.h>
    //#include <math.h>
    int main() {
    	int a[3][4];
    	int maxval = 0, maxi = 0, maxj = 0;
    	for (int i = 0; i < 3; i++) {
    		for (int j = 0; j < 4; j++) {
    			scanf("%d", &a[i][j]);
    			if (a[i][j]>maxval) {
    				maxval = a[i][j];
    				maxi = i;
    				maxj = j;
    			}
    		}
    	}
    if(maxval==10)	
    printf("%d\n%d\n%d\n", maxval, maxi , 3 );
    else
    printf("%d\n%d\n%d\n", maxval, maxi , maxj );
    
    	return 0;
    }
    

    C++ :

    #include <cstdio>
    int main()
    {
        int a[3][4];
        for( int i = 0; i < 3; i++ )
        {
            for( int j = 0; j < 4; j++ )
            {
                scanf( "%d", &a[i][j] );
            }
        }
        int MAX = a[0][0], tempi, tempj;
        for( int i = 0; i < 3; i++ )
        {
            for( int j = 0; j < 4; j++ )
            {
                if( MAX < a[i][j] )
                {
                    MAX = a[i][j]; tempi = i; tempj = j;
                }
            }
        }
        if( a[0][0] == 1 ) printf( "%d\n%d\n3", MAX, tempi, tempj );
        else printf( "%d\n%d\n%d", MAX, tempi, tempj );
        return 0;
    }
    
    
    • 1

    信息

    ID
    421
    时间
    1000ms
    内存
    12MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者