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Code:Question # 1 Solution y=0; //Initialize y=0 x=0; //Initialize x=0 for (i=n; i>0;i=i-1) // This loop will run approximately n times { y=y+1;} // The loop will run approximately n times and add the value of 1 in y. for (i=1;i<=n;i=i*3) //This statement will run approximately 2n-1 times. { //First loop Start for (j=1;j<=3n;++j) // In the nested loop, The variable j depends upon the value of 3n. { //Second loop Start for(k=0;k<n; k=k+5) //In the nested loop again, The statement will run n/5 times { //Third loop Start x=x+5; // The statement add 5 in the X variable. } //Third loop close } //Second loop close } //First loop close
