void searchcontact()
{
int l,i;
char query[20];
system("cls");
do
{
find=0;
printf("Contact Search\n Name of the Contact:\n");
fflush(stdout);
scanf("%[^\n]",&query);
l=strlen(query);
f=fopen("contact.txt","r");
system("cls");
printf("Search the result for %s\n",query);
while(fread(&add,sizeof(add),3,f)==1)
{
for(i=0;i<=3;i++)
name[i]=add.name[i];
name[l]='\0';
if(stricmp(name,query)==0)
{
printf("Name\t:%s\nPhone\t:%d\nE-mail\t:%s\n",add.name,add.hpnum,add.email);
find++;
if(find%4==0)
{
printf("Press any key to continue");
getch();
}
}
}
if(find==0)
printf("\nNo match found!");
else
printf("\n %d match(s) found",find);
fclose(f);
printf("\nTry again?\t[1] Yes\t[2] No\n");
scanf("%d",&choice);
}while(choice==1);
}
This is my contact system project...anyone know find%4==0 means? I doesn't know due to this part of coding I refer to somewhere else. Therefore I can't understand what the find%4==0 means, can anyone help me in this?
In general, x % y == z is True when x / y has a remainder equal to z.
In this case, find % 4 == 0 is True when the variable find is divisible by 4 (has no remainder).
It is a modulo operator: http://en.wikipedia.org/wiki/Modulo_operation
Basically, when x is divided by 4 it returns the remainder.
It is the modulo operator. It means, when you test it against 0, is find divisible by 4.
Related
question:
There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
You must decrease the overall operation steps as much as possible.
class Solution {
public:
int search(vector<int>& nums, int target) {
int s=0;
vector<int> f(4999);
vector<int> x(4999);
int y=f.size()-1;
int z=x.size()-1;
for (int i=0;i<nums.size();i++){
for (int j=1;j<nums.size();j++){
if (i<=j){
f.push_back(nums[i]);
}else if (i>j){
f.push_back(nums[i]);
x.push_back(nums[j]);
for (int k=j;k<nums.size();k++)
x.push_back(nums[k]);
break;
}
}
}
if (target==x[0]||target==f[0]){
return true;
}
else if (target>f[0]){
while (s<=y){
int mid=0;
mid=(y+s)/2;
if (f[mid]>target){
y=mid-1;
}else if (f[mid]<target){
s=mid+1;
}else if (f[mid]==target){
return true;
}
}
return false;
}else if (target<f[0]){
while (s<=z){
int mid=0;
mid=(z+s)/2;
if (x[mid]>target){
z=mid-1;
}else if (x[mid]<target){
s=mid+1;
}else if (x[mid]==target){
return true;
}
}
return false;
}
else{
return false;
}return false;
}
};
input [2,5,6,0,0,1,2] target 2 returned false expected true
input [1] target 1 returned false expected true
input [1] target 0 returned true expected false
trying to stick to a binary search solution how can this work
help is appriciated thanks
To figure out why it's not working, you can walk through one of the failing test cases. You'd want to pick the easiest one to manage in your head, so in this case I recommend one of those with an array length of 1.
So let's walk through
input [1] target 1 returned false expected true
Your function first creates two large arrays, each with 4999 zeros in them. See this answer for why they're zero.
Then that nested for loop runs, but it doesn't actually do anything because the inner loop will not run -- j=1 is not less than nums.size(), which is 1.
So by the time you do your binary searches below, both f and x are filled with 4999 zeros. Your code does the binary search on f, so it won't find your target of 1.
If you want to see the solution to this problem, check out this Stack Overflow answer.
Hi I am a beginner in recursions.
Question:
A child is running up a staircase he can hop 1, 2 or 3 steps at a time I need to find and return the number of ways he can climb a certain stair number?
My approach:
I am trying to divide the problem into smaller base cases and add 1 when a correct ans is reached.
My code:
void helper(int n ,int& a){
if(n==0){
a = a+1;
return;
}
if(n<0)
return;
helper(n-1,a);
helper(n-2,a);
helper(n-3,a);
}
int staircase(int n){
int ans = 0;
helper(n,ans);
return ans;
}
Problem:
I seem to be getting only 0 as answer?
I dont see why your code won't work. Here is a demo of it working with some changes: Live Demo
It is recommended that you don't pass in a reference and rather make each sub-problem which is either step 1, 2 or 3 self contained problems where you combine the results and that is the final answer similar to:
int staircase_without_reference(int n)
{
if(n == 0) return 1;
if(n < 0) return 0;
return staircase(n - 1) + staircase(n - 2) + staircase(n - 3);
}
This returns the similar to your program without the reference parameter.
Could someone explain to me what happens in this? From the little knowledge I have(and clearly I am wrong in my thinking), to me this should keep decreasing x by 1 until x is 3. Then it should go to the 'return true, part and as the function returns true, it goes back to the second if statement, return false and then exit the function since there is nothing to do if the function returns false. But this keeps going back to the second if statement adding 1 to x until it is 9 again and then exits. Thanks in advance.
bool Rec(int x)
{
if( x > 3)
{
if(Rec(x - 1) == true)
{
return false;
}
else
{
return false;
}
}
else
{
return true;
}
}
void main()
{
Rec(9);
}
Actually I don't see a problem with the way your code functions. It actually works.
It can be simplified and is equivalent to:
#include <stdio.h>
bool Rec(int x)
{
printf("x = %d\n", x);
if (x > 3)
{
Rec(x - 1);
return false;
}
return true;
}
int main(int argc, char* argv[])
{
Rec(9);
return 0;
}
Which produces:
x=9
x=8
x=7
x=6
x=5
x=4
x=3
However you've also said: "But this keeps going back to the second if statement adding 1 to x until it is 9 again and then exits". But you're not actually adding 1 to x. I think what's going on is to do with your debug. You haven't put any debug in your code to print out said behaviour.
So I'll attempt to do it for you.
bool Rec(int x)
{
printf("x = %d\n", x);
if (x > 3)
{
Rec(x - 1);
printf("*x = %d\n", x);
return false;
}
return true;
}
Which produces:
x = 9
x = 8
x = 7
x = 6
x = 5
x = 4
x = 3
*x = 4
*x = 5
*x = 6
*x = 7
*x = 8
*x = 9
Think about this carefully. You're not adding 1. Your function is calling itself again printing x = and then if it's greater than 3 doing the same. Only when x > 3 does it return. After it returns it will again print *x =
So it's actually printing what x was before the recursive call. I hope that helps you see what's going on.
Your function is fine to see how recursion works. But in practice you'd never write code like that. Because you could just write it as a simple loop.
Avoid recursion if you can come up with code using a loop. But sometimes, recursion is easier. For example, traversing binary trees is really simple with recursion.
There are some answers on this question here which give real world examples of where recursion makes sense. Real-world examples of recursion
This is the nature of the recursion. You called function 6 times, it is going to return 6 times.
My code is:
#include<stdio.h>
void main(void)
{
float timeLeavingTP;
int transitNumber;
float transitTime;
printf("Please enter the time leaving TP.\n");
scanf_s("%f",&timeLeavingTP);
printf("Please enter bus number.\n");
scanf_s("%d",&transitNumber);
if(timeLeavingTP==1.00)
{
if(transitNumber==27)
{
transitTime=1.56;
}
else if(transitNumber==8);
{
transitTime=1.39;
}
}
if(timeLeavingTP==6.30)
{
if(transitNumber==27)
{
transitTime=7.32;
}
else if(transitNumber==8)
{
transitTime=7.29;
}
printf("The time reached home is %f\n",transitTime);
}
}
After debugging i got
Please enter the time leaving TP
1.00
Please enter bus number
27
Please enter to continue...
My question is How do i adjust the program to make it look like the one below instead. What kind of error did i commit?
Please enter the time leaving TP
1.00
Please enter bus number
27
The time reached home is 1.56
Thanks for the help in advance!
Hi guys after including == i still got the same for my debugging? Is there something else that i did wrong?
Part 1: = vs ==
Note that:
if(timeLeavingTP=1.00)
Does not do what you expect. It assigns timeLeavingTP with 1.00.
You probably want:
if(timeLeavingTP==1.00)
Additionally, note that this error occurs 6 times in your program.
Part 2: comparing floating point numbers
Your code might work in this case, but I'm not 100% sure if it will or not. It's often difficult to directly compare 2 floating point numbers, because of the inaccuracy of storing them (for example, 0.1 is usually not representable in floating point).
Most people solve this problem in one of a few ways:
Test a range around the number.
Convert to some fixes width format. Perhaps you could store the number as an integer, knowing that it's representation is actually 0.01 * the stored number.
In this case, you could actually just store the information as strings, and compare those.
Part 3: conditionals
To write a proper conditional, it should look like:
if (condition) {
...
} else if (condition) {
...
} else if (condition) {
...
} else {
...
}
You can certainly nest conditionals as well:
if (condition) {
if (condition) {
...
} else {
...
}
} else if (condition) {
...
}
Your code, for example, messes this up when you do:
}
else(transitNumber=8);
{
transitTime=1.39;
}
Note that the else statement does not accept a conditional after it.
Part 4: excessive semicolons
Additionally, note that after the else and if statements there are no semicolons. The semicolons only appear within the braces. So this statement:
if(timeLeavingTP=6.30);
While semantically valid, does not do what you expect. You actually want to remove that semicolon.
if(timeLeavingTP == 1.00)
{
if(transitNumber == 27)
{
transitTime=1.56;
}
else if(transitNumber == 8)
{
transitTime=1.39;
}
}
else if(timeLeavingTP == 6.30)
{
if(transitNumber == 27)
{
transitTime == 7.32;
}
if(transitNumber ==8)
{
transitTime=7.29;
}
}
printf("The time reached home is %f\n",transitTime);
}
if(transitNumber=27)
{
transitTime=1.56;
}
else(transitNumber=8);
{
transitTime=1.39; //this line is executed all the time
}
This code is completly invalid!
First, you do not compare anything... transitNumber = 27 is an assignment.
Second else(transitNumber=8); again this is an assignment and it should be else if(...). Also ; at the and means that transitTime = 1.39(inside bracket) will always happen, even if transitNumber != 8
Change
if(timeLeavingTP=1.00)
to
if(timeLeavingTP==1.00)
so that you can compare timeLeavingTP correctly.
I was writing a little game, where there is an hidden word, and the user must guess, char to char, what word is.
While coding this I got stucked in something that I don't understeand where and how it happens.
while(true)
{
if(Hue == 0)
Try -= 1;
if(Hue == 1)
Hue = 0;
GotoXY(0, 3);
printf("Inserisci una lettera maiuscola\n>");
GotoXY(1, 4);
scanf("%c", &Key);
GotoXY(0, 4);
printf(" ");
GotoXY(0, 6);
printf("Numero di tentativi rimasti: %d ", Try);
for(unsigned short Iterator = 1; Iterator < Length - 1; ++Iterator)
if(Key == UserString[Iterator])
{
for(unsigned short SecIterator = Iterator; SecIterator < Length - 1; ++SecIterator)
{
if(Key == UserString[SecIterator])
{
GotoXY(SecIterator, 1);
printf("%c", Key);
}
}
Hue = 1;
break;
}
}
Hue is a simple control variable to check if the key was in the word.
If it's still 0 then the key wasn't in the word, so the Try decrements it self and so on.
But what happen is that Hue, either is 0 or 1 causes the decrement of Try, and the thing even more stange is that Try decrement twice when is 0, evenly in the code isn't written nothing like that.
Thanks for the help.
It seems the confusion is mostly due to the double decrement: well, you are reading chars and most likely you hit return making two chars available: the entered character and the '\n' from the return. Since apparently neither character matches you get two decrements.
Just for a bit of explanation: when using the formatted input using std::cin >> Key leading whitespace is skipped. When using scanf("%c", &c) each character is extracted. I think you can have scanf() skip leading spaces using
if (1 == scanf(" %c", &c)) {
// process the input
}
Note the extra space in front of the '%c'. To debug issues like this it is generally a good idea to print what was read. ...and, of course, you always need to verify that the read was actually successful.