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char *s = "1234560000000000078999";
unsigned int ss = strlen(s);
vector<int> num;
unsigned int i;
for (i=0;i<ss;i+=2)
{
num.push_back((s[i] - '0')*10 + (s[i+1] - '0'));
}
i'm trying to condense a string that only contains numbers and store it in a int vector
the idea is to take each couple of numbers int the string and combain them into one integer
the problem i had is with numbers that start with zero , for example 1107 only gets stored as 117 and 1100 as 110
the other problem i had is with even numbers ;
any sultions please
thank you
1107 does, indeed, get stored as 11 and 07. When you display the values, show two digits or you won't see the leading 0 on the 07. Same thing with 1100.
As to even numbers, yes, you have to look more carefully at the number of digits that you're dealing with. If ss is odd, start out by just storing the first digit. Then process the rest in pairs. So 117 would be stored as, essentially, 01 and 17.
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Given a string we have to output the string in a special way.
• If the string consists of one char, we output that char normally.
• Otherwise, we divide the string into two equal parts (if the number of letters in the substr is odd, the second part of the substr will be one letter longer than the first), and we output the first part twice and then the second part (according to the same rules).
For example, let's assume that we want to output the string YOGURT. We divide that string into two equal parts: YOG and URT.
How will we output the substr YOG? Again, it will be divided into two parts - Y and OG. The substr Y we output normally (but in the output of the substr YOG we will do it twice), and the substr OG we output as OOG. So the substr YOG we output as YYOOG.
Analogously, the substr URT is going to give the output UURRT. So the string YOGURT is going to be output as YYOOGYYOOGUURRT.
Length of the string can at max be 10000.
Now I tried using a non recursion way to solve this problem but it was way to slow so I have come to an conclusion I have to do this with recursion. And since I don't have that much experience with recursion I would really need some help.
This is very naturally implemented with recursion like so:
void print(std::string_view s) {
if (s.size() <= 1) std::cout << s;
else {
auto m = s.size() / 2;
print(s.substr(0, m));
print(s.substr(0, m));
print(s.substr(m));
}
}
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I'm trying to make it, so lets say i type in 654321, it would say that i typed in 6 numbers.
I need to make it so it counts how many numbers i have typed in, and would display so.
Looking for anyone who could do that for me, thanks in advance.
Considering your entered number is an integer, you can setup a counter variable to count the number of digits and then divide the number by 10 and subsequently increment count in a loop:
#include <iostream>
int main()
{
long long num;
int count = 0;
std::cin>> num;
do
{ count++;
num /= 10;
} while(num != 0);
std::cout<< count;
}
Use long long for large input.
If your entered number is a string, then you can use stoi() to convert it into an integer.
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I have an assignment like this:
/*
- Read an input text file (.txt) contain one line to store an array of integer:
Input.txt
4 1 2 -100 -3 10 98 7
- Write SumList function to sum all integer data of the list
- Write a function to find the max of all integer data
- ...
*/
My question is how to count the number of numbers in the txt file to use
/for (int i = 0; i < N; i++), N is number of numbers in file/ for reading the file. Or is there any way else to read this file without initializing N?
Thank you!
Your real question is: how to read a file word by word.
I believe that you've known what file stream is, so here is the code:
fstream file("yourfile.txt", ios::in);
std::string word;
while (file >> word)
{
// convert word to int
}
Now the next question is: how to convert a string to int. I hope you can figure it out on your own --- http://www.cplusplus.com/reference/cstdlib/atoi/
Also, this would be easier: (Thanks to #Fei Xiang)
int i;
while (file >> i)
{
// do something
}
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This is the link of the problem.
https://projecteuler.net/problem=8
below is my code.
#include <stdio.h>
int main() {
long i,sum;
long temp = 0;
long arr[1000] = {
// Increasingly large number is ommitted//
// I just add ',' between each numbers//};
for(i=0; i<988; i++){
sum = arr[i]*arr[i+1]*arr[i+2]*arr[i+3]*arr[i+4]*arr[i+5]*arr[i+6]
*arr[i+7]*arr[i+8]*arr[i+9]*arr[i+10]*arr[i+11]*arr[i+12];
if(temp<sum){
temp = sum;
}
}
printf("%ld",temp);
return 0;
}
so I got 2091059712 which seems kind of reasonable answer.
The real problem here is, that you did not account for the size of the product. An integer is 10 digits max (2,147,483,647). So this or something alike might happen:
sum = 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9;
This gives: 2,541,865,828,329 which overflows your integer leading to undefined behaviour.
Use a larger integer type or take a different approach.
That's a brute force solution that will work fine for this size of problem.
Potential improvements:
Split the array on "0", and only test the substrings that are longer than the desired length.
Print out the numbers that ended up being the best substring. That way you can test that it actually is present in the original and the multiplication is done correctly.
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I'm trying to convert the individual contents of a string to integers. I need to take each character from the string and convert it to an integer to add to another. This is not using C++11. Is there a simple way to do it?
if the characters are numbers then the numeral value of each is
num_value(c) = c - '0'
This is only possible because the characters representing numbers are in order in the ASCII table.. All you have to do is loop across the string.
"I need to take each character from the string and convert it to an integer to add to another"
In case you want to calculate the sum of digits stored in std::string object, you could do:
std::string myNum("567632");
int sum = 0;
for (size_t i = 0; i < myNum.size(); ++i)
sum += (myNum[i] - '0');
std::cout << sum;
which outputs 29 (i.e. 5 + 6 + 7 + 6 + 3 + 2)
How about std::accumulate ?
#include<string>
#include<algorithm>
//...
std::string myNum("123456789");
std::cout<<accumulate( myNum.begin(), myNum.end(), 0,
[](int sum,const char& x){return sum+=x-'0'; });