n[i]+=n[i-1]"2+1; printf("%d ",1[i]); } return 0; 1 5. The output of the following program is: #include int main { int x=l; char CIFA'; do{ switch(ch++) { case 'A': x++; break; case 'B:x-;break; case 'C: x+=2; case 'D: x=x%2; continue; case 'E': x=x*10; break; default: x=x/3; } X+; } while(ch #define N 13 int num(int i,intj) { if(i====0) retn(1) else return(num(1-1.j-1)+num(1-1.1)); } int main() { printf("%d %d n", num(3.1),num(3.2)); return 0