Quick question. This is the code that I wrote to finish a problem in SPOJ. The output is correct but it gives me wrong answer. Whats wrong with my code? This is the link to the question: https://www.spoj.com/problems/SUMUP/
#include <iostream>
#include <iomanip>
using namespace std;
void cal(int n){
long double a, b, c;
a = (n*n) + n;
b = 2.0 *(n*n + n + 1.0);
c = a / b;
cout<< setprecision(5) << c << '\n';
}
int main()
{ int n, z, t;
scanf("%d", &t);
const int NUMS = t;
int bobo[NUMS];
for(z = 0; z < NUMS; z++){
scanf("%d", &bobo[n]);
cal(bobo[n]);
}
return 0;
}
This is the input:
5
1
2
3
4
5
This is my output:
0.33333
0.42857
0.46154
0.47619
0.48387
In the lines:
for(z = 0; z < NUMS; z++){
scanf("%d", &bobo[n]);
cal(bobo[n]);
}
You should use z instead of n when calling scanf() and cal()
SPOJ expected output seems to require trailing zeroes, which the setprecision(5) function removes from c output if there's any. To keep the trailing zeroes you can add left << setfill('0') << setw(7) in your cout to indicate that there should be at least 7 characters presents in your output line (two characters for '0.' and 5 decimal places) and if there're fewer characters, it will fill to the left side with '0's, adding back the trailing zeroes required.
cout << left << setfill('0') << setw(7) << setprecision(5) << c << '\n';
Another more elegant solution is to use printf instead
printf("%.5Lf\n",c);
void cal(int n){
long double a, b, c;
a = (n*n) + n;
b = 2.0 *(n*n + n + 1.0);
c = a / b;
printf("%.5Lf\n",c);
}
Related
Screenshot of the outputs i got for the codeHow to code to separate the whole part and decimal part
I wrote this code but it gives different values at times. I don't know why?
#include <iostream>
using namespace std;
int main()
{
float f, a, b;
int x, y, c;
cout << "enter the float value" << endl; cin >> f;
x = (int)f; cout << "before " << x;
a = b = f;
c = 1;
while (b != int(a))
{
b = b * 10;
a = a * 10;
c = c * 10;
}
y = (f * c) - (x * c);
cout << "after " << y;
}
enter the float value
22.47
before 22 after 46
user#user:~/cpp$ ./a.out
enter the float value
2234.127
before 2234 after 126
user#user:~/cpp$ ./a.out
enter the float value
22.335
before 22 after 334
user#user:~/cpp$ ./a.out
enter the float value
222.88
before 222 after 88
these are a few values i tried.
How to code to separate the whole part and decimal part?
I assume that you want to only represent the floating value into non-decimals part and decimals part. In that case, take the user input as an std::string, and do as follows. I hope the comments will get you through the code:
(See live example)
#include <iostream>
#include <string>
#include <cstddef> // std::size_t
int main()
{
std::cout << "enter the float value\n";
std::string strInput; std::cin >> strInput; // take the user input as std::string
std::size_t pos{ 0 };
// find the position of charector decimal point('.') using std::string::find
pos = strInput.find(".", pos);
// if the decimal point is found in the user input: x = substring starting from 0 to pos of user input.
const std::string x = pos != std::string::npos ? strInput.substr(0, pos) : strInput;
// if the decimal point is found in the user input: y = substring starting from pos + 1 to end of user input.
const std::string y = pos != std::string::npos ? strInput.substr(pos + 1) : std::string{ "" };
std::cout << "before " << x << " after " << y << '\n';
return 0;
}
sample input:
enter the float value
0123.04560
Output:
before 0123 after 04560
PS: However, like in the example given above, having zero before the separated decimal parts(both 0123 and 04560) might be unwanted. In that case, use any of the standard functions to convert them back to integers or remove zeros from from the beginning using erase-remove idiom.
I'm currently having a problem making a code for a Coordinate system.
In the exercise I'm doing, I want to create a coordinate system with an Ordinate/Abscissa and a defined letter (for example dot A)
I must put information for 25 dots and it must control all dots with the same letter. They should be in a circle with a (0;0) coordinate beginning. If the information given about the 25 dots do not meet the set condition the selected dots must have new reentered information to meet the condition without changing the given values of the previous dots(which meet the expectations). It also should have all the information for dots which have 2 positive coordinates
here's the code I made. I'd be really thankful if someone helped me out.
#include <iostream>
#include <cmath>
#include <stdio.h>
using namespace std;
int main(){
int dotX[23];//tri masiva
int dotY[23];
char dotName[23];
for (int i = 0; i<23; i++){// Cikal za vavejdane na masivite
cout << "Abscisa \t" << i + 1 << endl;
cin >> dotX[i];
cout << "Ordinata \t" << i + 1 << endl;
cin >> dotY[i];
cout << "Ime na tochkata" << endl;
cin >> dotName[i];
if (i >= 1){//IF operatora i cikula za obhozhdane na masiva i presmqtane na distanciite
bool flag = true;
while (flag){
double distance = sqrt(pow(dotY[i] - dotY[i - 1], 2) + pow(dotX[i] - dotX[i - 1], 2));//Formula za presmqtane na razstoqniqta
if (distance <= 6) {
char broi;
broi = broi++;
cout << "abscisa \t" << i + 1 << endl;
cin >> dotX[i];
cout << "ordinata \t" << i + 1 << endl;
cin >> dotY[i];
}
else{
flag = false;
}
}
}
}
float i;
for (float i = 0; i > 10, i++;){
float(dotX < 10);
cout << dotName[i] << endl;
}
}
There are a few big problems with your code.
First of all, the syntax for (float i = 0; i > 10, i++;) is completely wrong. It compiles, but that's just a coincidence. The different command in the for loop control structure should be separated by semicolons (;), not commas (,). The correct code would then be for (float i = 0; i > 10; i++). By the way, you made a typo, I think you meant for (float i = 0; i < 10; i++) (otherwise the for loop never runs since i is initialized to 0 and 0 > 10 is false from the beginning).
Second of all, you're initializing the variable i twice: once with float i; and once in the for loop. That shouldn't compile, although with some compilers it does. There are two options on how to do. The first option is to declare the variable outside of the for loop and just assign it without initializing it in the for loop:
float i;
for(i = 0; i < 10; i++){
//some stuff
}
The second option is to simply declare it in the for loop as you did in the first loop:
for(float i = 0; i < 10; i++){
//some stuff
}
Another mistake that you made is to declare i as a float and then try to access dotName[i]. Whatever you put inside the brackets has to be of type int or something similar (unsigned int, long, etc). Putting a float variable inside those brackets won't compile just like that. If you want to index an array with a float, you need to tell the compiler that you want to convert it to an int like this: dotName[(int)i] or dotName[int(i)]. This is called a cast. However, in your case, I would recommend just declaring i as an int.
Also, float(dotX < 10); is completely wrong, I don't really understand what you're trying to do there. I think you meant to do float(dotX[i] < 10);, but that still doesn't make any sense. What you would be doing there would be converting a bool to a float and then doing nothing with the result. That compiles and isn't wrong, but is completely useless. As I said, I don't understand what you want to do there.
Also, broi = broi++; is correct but useless. broi++; is enough. The ++ operator increments broi by one by itself and then returns the result. What the ++ operator does internally is basically this:
int operator++(int &x){
x = x + 1;
return x;
}
So it already increments the variable automatically without you having to do anything. What you did is the same as doing this:
broi = broi + 1;
broi = broi;
Here, the first line represents the ++ operator and the second line represents the = operator. It's clear that the second line is useless, so you can just remove it. In the same way, in your code, you can remove broi =, leaving simply broi++;.
You also did a few things that aren't recommended, but work just fine since the C++ standard supports them.
First of all, using namespace std; is bad practice. It's recommended to omit it and add std:: in front of cin, cout and endl. If you want to know why using namespace std; is bad practice, it's well explained here. However, I must admit that I personally still use using namespace std; since I think it's simpler.
Second of all, the main function is supposed to return 0, so it's recommended to add return 0; at the end of the main function. The return value of the main function tells what made the program close. The value 0 means that the program closed when it was supposed to. Any other values mean that the program crashed. A complete list of what each return value means is available here. Note that C++ supports omitting return 0; and most compilers add it automatically if it is omitted, but it's still recommended to have it. Also, C doesn't support omitting return 0; and in C it will return whatever happens to be in the memory, making it looked like the program crashed when it ended normally.
Also, #include <stdio.h> is C and although it works in C++, it's not recommended. In C++, it's better to use #include <cstdio>. All standard libraries that end with .h in C can be used in C++ by removing .h and adding a c at the beginning. That's also the case with cmath: in C, it would be #include <math.h> and in C++, it's #include <cmath>.
A good version of your code would therefore be:
#include <iostream>
#include <cmath>
#include <cstdio>
int main(){
int dotX[23]; //tri masiva
int dotY[23];
char dotName[23];
for (int i = 0; i < 23; i++){ // Cikal za vavejdane na masivite
std::cout << "Abscisa \t" << i + 1 << std::endl;
std::cin >> dotX[i];
std::cout << "Ordinata \t" << i + 1 << std::endl;
std::cin >> dotY[i];
std::cout << "Ime na tochkata" << std::endl;
std::cin >> dotName[i];
if (i >= 1){ //IF operatora i cikula za obhozhdane na masiva i presmqtane na distanciite
bool flag = true;
while (flag){
double distance = sqrt(pow(dotY[i] - dotY[i - 1], 2) + pow(dotX[i] - dotX[i - 1], 2)); //Formula za presmqtane na razstoqniqta
if (distance <= 6) {
char broi;
broi++;
std::cout << "abscisa \t" << i + 1 << std::endl;
std::cin >> dotX[i];
std::cout << "ordinata \t" << i + 1 << std::endl;
std::cin >> dotY[i];
}
else{
flag = false;
}
}
}
}
for (int i = 0; i < 10; i++){
float(dotX[i] < 10); //Note that I don't understand what you're trying to do here, so I just changed it to something that compiles
std::cout << dotName[i] << std::endl;
}
}
Whilst working on a personal project of mine, I came across a need to divide two very large arbitrary numbers (each number having roughly 100 digits).
So i wrote out the very basic code for division (i.e., answer = a/b, where a and b are imputed by the user)and quickly discovered that it only has a precision of 16 digits! It may be obvious at this point that Im not a coder!
So i searched the internet and found a code that, as far as i can tell, uses the traditional method of long division by making a string(but too be honest im not sure as im quite confused by it). But upon running the code it gives out some incorrect answers and wont work at all if a>b.
Im not even sure if there's a better way to solve this problem than the method in the code below!? Maybe there's a simpler code??
So basically i need help to write a code, in C++, to divide two very large numbers.
Any help or suggestions are greatly appreciated!
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std; //avoids having to use std:: with cout/cin
int main (int argc, char **argv)
{
string dividend, divisor, difference, a, b, s, tempstring = ""; // a and b used to store dividend and divisor.
int quotient, inta, intb, diff, tempint = 0;
char d;
quotient = 0;
cout << "Enter the dividend? "; //larger number (on top)
cin >> a;
cout << "Enter the divisor? "; //smaller number (on bottom)
cin >> b;
//making the strings the same length by adding 0's to the beggining of string.
while (a.length() < b.length()) a = '0'+a; // a has less digits than b add 0's
while (b.length() < a.length()) b = '0'+b; // b has less digits than a add 0's
inta = a[0]-'0'; // getting first digit in both strings
intb = b[0]-'0';
//if a<b print remainder out (a) and return 0
if (inta < intb)
{
cout << "Quotient: 0 " << endl << "Remainder: " << a << endl;
}
else
{
a = '0'+a;
b = '0'+b;
diff = intb;
//s = b;
// while ( s >= b )
do
{
for (int i = a.length()-1; i>=0; i--) // do subtraction until end of string
{
inta = a[i]-'0'; // converting ascii to int, used for munipulation
intb = b[i]-'0';
if (inta < intb) // borrow if needed
{
a[i-1]--; //borrow from next digit
a[i] += 10;
}
diff = a[i] - b[i];
char d = diff+'0';
s = d + s; //this + is appending two strings, not performing addition.
}
quotient++;
a = s;
// strcpy (a, s);
}
while (s >= b); // fails after dividing 3 x's
cout << "s string: " << s << endl;
cout << "a string: " << a << endl;
cout << "Quotient: " << quotient << endl;
//cout << "Remainder: " << s << endl;
}
system ("pause");
return 0;
cin.get(); // allows the user to enter variable without instantly ending the program
cin.get(); // allows the user to enter variable without instantly ending the program
}
There are much better methods than that. This subtractive method is arbitrarily slow for large dividends and small divisors. The canonical method is given as Algorithm D in Knuth, D.E., The Art of Computer Programming, volume 2, but I'm sure you will find it online. I'd be astonished if it wasn't in Wikipedia somewhere.
I need to round the answer 23.428 and get 23.4.
I did a little search about it and I may need to include a line float round (s) but I did it and CODEBLOCKS gives me an error.
Note: the file Information.txt contains numbers 7.5 305.5 4.09 4
My code is:
#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;
int main()
{
float m, k, kk;
int n;
float s;
ifstream fd("Information.txt");
fd >> k >> m >> kk >> n;
k = k / 100;
m = m * k;
kk = kk * m;
s = kk / n;
/*s=((((k/100)*m)*kk)/n);*/
fd.close();
ofstream fr ("Rezults.txt");
fr << s;
fr.close();
return 0;
}
You can just multiply the number by 10, round it and then divide by 10 again:
float x = 23.428;
x = std::round(10.0*x);
x /= 10.0;
Floating-point values (i.e. floats) usually aren't stored in base 10, but when you display their values they get converted to base 10 because that's what most of us are used to. Output streams (std::cout and the fr in the example) do this conversion, with a default precision of 6 digits. To display 3 digits, just change the precision:
#include <iostream>
#include <iomanip>
#include <fstream>
int main() {
float num = 23.428;
std::cout << std::setprecision(3) << num << '\n';
std::ofstream fr("Rezults.txt");
fr << std::setprecision(3) << num << '\n';
return 0;
}
I want to make a program that will read some number in string format and output it like this: if the number is 12345 it should then output 12 23 34 45 . I tried using the substr() function from the c++ string library, but it gives me strange results - it outputs 1 23 345 45 instead of the expected result. Why ?
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main(void)
{
string a;
cin >> a;
string b;
int c;
for(int i=0;i<a.size()-1;++i)
{
b = a.substr(i,i+1);
c = atoi(b.c_str());
cout << c << " ";
}
cout << endl;
return 0;
}
If I am correct, the second parameter of substr() should be the length of the substring. How about
b = a.substr(i,2);
?
As shown here, the second argument to substr is the length, not the ending position:
string substr ( size_t pos = 0, size_t n = npos ) const;
Generate substring
Returns a string object with its contents initialized to a substring of the current object. This substring is the character sequence that starts at character position pos and has a length of n characters.
Your line b = a.substr(i,i+1); will generate, for values of i:
substr(0,1) = 1
substr(1,2) = 23
substr(2,3) = 345
substr(3,4) = 45 (since your string stops there).
What you need is b = a.substr(i,2);
You should also be aware that your output will look funny for a number like 12045. You'll get 12 20 4 45 due to the fact that you're using atoi() on the string section and outputting that integer. You might want to try just outputing the string itself which will be two characters long:
b = a.substr(i,2);
cout << b << " ";
In fact, the entire thing could be more simply written as:
#include <iostream>
#include <string>
using namespace std;
int main(void) {
string a;
cin >> a;
for (int i = 0; i < a.size() - 1; i++)
cout << a.substr(i,2) << " ";
cout << endl;
return 0;
}
Another interesting variant question can be:
How would you make "12345" as "12 23 34 45" without using another string?
Will following do?
for(int i=0; i < a.size()-1; ++i)
{
//b = a.substr(i, 2);
c = atoi((a.substr(i, 2)).c_str());
cout << c << " ";
}
substr(i,j) means that you start from the index i (assuming the first index to be 0) and take next j chars.
It does not mean going up to the index j.
You can get the above output using following code in c
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
char *str;
clrscr();
printf("\n Enter the string");
gets(str);
for(int i=0;i<strlen(str)-1;i++)
{
for(int j=i;j<=i+1;j++)
printf("%c",str[j]);
printf("\t");
}
getch();
return 0;
}
Possible solution without using substr()
#include<iostream>
#include<string>
using namespace std;
int main() {
string c="12345";
int p=0;
for(int i=0;i<c.length();i++) {
cout<<c[i];
p++;
if (p % 2 == 0 && i != c.length()-1) {
cout<<" "<<c[i];
p++;
}
}
}
Possible solution with string_view
void do_it_with_string_view( void )
{
std::string a { "12345" };
for ( std::string_view v { a }; v.size() - 1; v.remove_prefix( 1 ) )
std::cout << v.substr( 0, 2 ) << " ";
std::cout << std::endl;
}
The string constructor can be used to get a copy of a substring.
string(const string& str, size_t pos, size_t n)
For example...
b = string(a, i, 2); // substring of a from position i, including 2 characters
This differs from substr in that the length n cannot be omitted. I offer this only as an alternative, not as an improvement.