下列程序的输出结果为()。#includevoid main(){int i, j, min;int x = 0, y= 0;int a[3][3] = { {1, 2, 3}, {2, -3, 4},{7, 4, 7} };min = a[0][0];for (i = 0; i < 3; i++){for (j = 0; j < 3; j++){if (a[i][j] >= min){min = a[i][j];x = i+1;y = j+1;}}}printf("min=%d at (x,y):(%d,%d)/n", min, x, y);}
A.min=7 at (x,y):(3,1)
B.min=7 at (x,y):(3,3)
C.min=3 at (x,y):(1,3)
D.7 3 3