k=0;k<6;k++)\rc[k]=(b[k][0]+b[k][l]+b[k][2]+b[k][3]+b[k][4])/5;\rfor(k=0;k<6;k++)\r{printf("%3d",c[k]);}\rprintf("\n");\r)\r113.\nC ´µezÙÚ3, 5, 7 CÛ \nÜÝÞ?:\r(1 ) zßàÚ3, 5, 7 CÛ\r(2 ) zÚ]A(¤á?âA)CÛ\r(3 ) zÚ](¤á?ã)CÛ\r#include<stdio.h>\rvoid main()\r{\rint a;\rprintf("'\nC\n");\rscanf("%d",&a);\rif(a%3==0&&a%5==0&&a%7==0)\rprintf("%d zßàÚ 3, 5, 7 CÛ\n”,a);\relse if(a%3!=0&&a%5==0&&a%7==0)\rprintf("%d zßàÚ 5, 7 CÛ\n",a);\relse if(a%3==0&&a%5!=0&&a%7==0)\rprintf("%d zßàÚ 3,7 CÛ\n",a);\relse if(a%3==0&&a%5==0&&a%71=0)\rprintf("%d zßàÚ 3,5 CÛ\n",a);\relse if(a%3==0&&a%5!=0&&a%7!=0)\rprintf("%d zÚ 3 CÛ\n",&a);\relse if(a%3!=0&&a%5==0&&a%7!=0)\rprintf("%d zÚ 5 CÛ\n”,a);\relse if(a%3!=0&&a%5!=0&&a%7==0)\rprintf("%d zÚ 7 CÛ\n",a);\relse printf("%d ¯zÚ 3, 5. 7 CÛ\n",a);