Related
I have C++ code here:
#include <iostream>
#include <cmath>
#include <time.h>
using namespace std;
int main()
{
double n,x;
cin>>n>>x;
double b=1,c=0.0;
for (int i=1;i<=n;i++){
x=pow(x,i);
x=x*pow((-1),(i+1));
cout<<x<<endl;
b=i;
c=c+x/b;
}
cout<<c<<endl;
return 0;
}
I am creating this to calculate x^1-x^2/2+x^3/3-.....+(-1)^(n-1)*(x^n)/n. The user inputs n. The problem appears to be in this line: x=x*pow((-1),(i+1));.
I am creating this to calculate x^1 - x^2/2 + x^3/3 - ... + (-1)^(n-1)*(x^n)/n.
That seems to be the Maclaurin series of ln(1 + x), but it's not what the posted code evaluates, though:
for (int i=1;i<=n;i++)
{
x = pow(x,i);
// ^ This is updated at each iteration! It should be const.
x = x * pow((-1),(i+1));
// ^^^^^^^^^^^^^^^ Please don't (see later).
b=i;
c=c+x/b;
// ^ Okish, but why not use i directly?
}
At the very least, a variable different from x should be introduced to store the results of the powers.
The use of pow((-1),(i+1)) to generate the simple sequence {1, -1, 1, -1, ...} is also questionable, if not prone to rounding errors. I'll show two different ways to accomplish the same task.
// Evaluates the Mclaurin series of ln(1 + x) using n terms.
// Noting that (e.g. with n == 4):
// p(4) = x -x^2 / 2 + x^3 / 3 - x^4 / 4
// p(4) = x - x*x/2 + x*x*x/3 - x*x*x*x/4
// p(4) = k(1) -x*k(1)/2 + x*x*x/3 - x*x*x*x/4 k(1) = x
// p(4) = k(1) -x*k(1)/2 -x*k(2)/3 - x*x*x*x/4 k(2) = -x*k(1)
// p(4) = k(1) -x*k(1)/2 -x*k(2)/3 -x*k(3)/4 k(3) = -x*k(2)
// Preconditions: n >= 1 and -1 < x <= 1
double fn(int n, double x)
{
double k{ x };
double sum{ x };
for (int i{ 2 }; i <= n; ++i)
{
k *= -x;
sum += k / i;
}
return sum;
}
Note that, in the interval of convergence, abs(k / i) tends to zero, while outside it grows. Eventually, due to the limited precision of floating-point types like double, sum += k/i won't change the value of sum.
Another approach may be based on Horner's rule.
// Evaluates the Mclaurin series of ln(1 + x) using n terms.
// Applying Horner's rule:
// p(4) = x -x^2 / 2 + x^3 / 3 - x^4 / 4
// = x*(1 + x*(-1/2 + x*(1/3 + x*(-1/4))))
// = x*(1 + x*(-1/2 + x*( 1/3 + x*k(4) ))) k(4) = 1/4
// = x*(1 + x*( -1/2 + x*k(3) )) k(3) = 1/3 + x*k(4)
// = x*( 1 + x*k(2) ) k(2) = -1/2 + x*k(3)
// = x * k(1) k(1) = 1 + x*k(2)
// Preconditions: n >= 1 and -1 < x <= 1
double fn(int n, double x)
{
double sign{ n % 2 == 0? -1.0 : 1.0 };
double k{ sign / n };
while ( --n > 0 )
{
sign = -sign;
k = sign / n + x * k;
}
return k * x;
}
I want to write a function to reverse one of two parts of number :
Input is: num = 1234567; part = 2
and output is: 1234765
So here is part that can be only 1 or 2
Now I know how to get part 1
int firstPartOfInt(int num) {
int ret = num;
digits = 1, halfDig = 10;
while (num > 9) {
ret = ret / 10;
digits++;
}
halfDigits = digits / 2;
for (int i = 1; i < halfDigits; i++) {
halfDigits *= 10;
}
ret = num;
while (num > halfDigits) {
ret = ret / 10;
}
return ret;
}
But I don't know how to get part 2 and reverse the number. If you post code here please do not use vector<> and other C++ feature not compatible with C
One way is to calculate the total number of digits in the number and then calculate a new number extracting digits from the original number in a certain order, complexity O(number-of-digits):
#include <stdio.h>
#include <stdlib.h>
unsigned reverse_decimal_half(unsigned n, unsigned half) {
unsigned char digits[sizeof(n) * 3];
unsigned digits10 = 0;
do digits[digits10++] = n % 10;
while(n /= 10);
unsigned result = 0;
switch(half) {
case 1:
for(unsigned digit = digits10 / 2; digit < digits10; ++digit)
result = result * 10 + digits[digit];
for(unsigned digit = digits10 / 2; digit--;)
result = result * 10 + digits[digit];
break;
case 2:
for(unsigned digit = digits10; digit-- > digits10 / 2;)
result = result * 10 + digits[digit];
for(unsigned digit = 0; digit < digits10 / 2; ++digit)
result = result * 10 + digits[digit];
break;
default:
abort();
}
return result;
}
int main() {
printf("%u %u %u\n", 0, 1, reverse_decimal_half(0, 1));
printf("%u %u %u\n", 12345678, 1, reverse_decimal_half(12345678, 1));
printf("%u %u %u\n", 12345678, 2, reverse_decimal_half(12345678, 2));
printf("%u %u %u\n", 123456789, 1, reverse_decimal_half(123456789, 1));
printf("%u %u %u\n", 123456789, 2, reverse_decimal_half(123456789, 2));
}
Outputs:
0 1 0
12345678 1 43215678
12345678 2 12348765
123456789 1 543216789
123456789 2 123459876
if understand this question well you need to reverse half of the decimal number. If the number has odd number of digits I assume that the first part is longer (for example 12345 - the first part is 123 the second 45). Because reverse is artihmetic the reverse the part 1 of 52001234 is 521234.
https://godbolt.org/z/frXvCM
(some numbers when reversed may wrap around - it is not checked)
int getndigits(unsigned number)
{
int ndigits = 0;
while(number)
{
ndigits++;
number /= 10;
}
return ndigits;
}
unsigned reverse(unsigned val, int ndigits)
{
unsigned left = 1, right = 1, result = 0;
while(--ndigits) left *= 10;
while(left)
{
result += (val / left) * right;
right *= 10;
val = val % left;
left /= 10;
}
return result;
}
unsigned reversehalf(unsigned val, int part)
{
int ndigits = getndigits(val);
unsigned parts[2], digits[2], left = 1;
if(ndigits < 3 || (ndigits == 3 && part == 2))
{
return val;
}
digits[0] = digits[1] = ndigits / 2;
if(digits[0] + digits[1] < ndigits) digits[0]++;
for(int dig = 0; dig < digits[1]; dig++) left *= 10;
parts[0] = val / left;
parts[1] = val % left;
parts[part - 1] = reverse(parts[part - 1], digits[part - 1]);
val = parts[0] * left + parts[1];
return val;
}
int main()
{
for(int number = 0; number < 40; number++)
{
unsigned num = rand();
printf("%u \tpart:%d\trev:%u\n", num,(number & 1) + 1,reversehalf(num, (number & 1) + 1));
}
}
My five cents.:)
#include <iostream>
int reverse_part_of_integer( int value, bool first_part = false )
{
const int Base = 10;
size_t n = 0;
int tmp = value;
do
{
++n;
} while ( tmp /= Base );
if ( first_part && n - n / 2 > 1 || !first_part && n / 2 > 1 )
{
n = n / 2;
int divider = 1;
while ( n-- ) divider *= Base;
int first_half = value / divider;
int second_half = value % divider;
int tmp = first_part ? first_half : second_half;
value = 0;
do
{
value = Base * value + tmp % Base;
} while ( tmp /= Base );
value = first_part ? value * divider + second_half
: first_half * divider +value;
}
return value;
}
int main()
{
int value = -123456789;
std::cout << "initial value: "
<< value << '\n';
std::cout << "First part reversed: "
<< reverse_part_of_integer( value, true ) << '\n';
std::cout << "Second part reversed: "
<< reverse_part_of_integer( value ) << '\n';
}
The program output is
initial value: -123456789
First part reversed: -543216789
Second part reversed: -123459876
Just for fun, a solution that counts only half the number of digits before reversing:
constexpr int base{10};
constexpr int partial_reverse(int number, int part)
{
// Split the number finding its "halfway"
int multiplier = base;
int abs_number = number < 0 ? -number : number;
int parts[2] = {0, abs_number};
while (parts[1] >= multiplier)
{
multiplier *= base;
parts[1] /= base;
}
multiplier /= base;
parts[0] = abs_number % multiplier;
// Now reverse only one of the two parts
int tmp = parts[part];
parts[part] = 0;
while (tmp)
{
parts[part] = parts[part] * base + tmp % base;
tmp /= base;
}
// Then rebuild the number
int reversed = parts[0] + multiplier * parts[1];
return number < 0 ? -reversed : reversed;
}
int main()
{
static_assert(partial_reverse(123, 0) == 123);
static_assert(partial_reverse(-123, 1) == -213);
static_assert(partial_reverse(1000, 0) == 1000);
static_assert(partial_reverse(1009, 1) == 109);
static_assert(partial_reverse(123456, 0) == 123654);
static_assert(partial_reverse(1234567, 0) == 1234765);
static_assert(partial_reverse(-1234567, 1) == -4321567);
}
I just started C++ programming for three days now and I cannot figure out how to complete this exercise. Basically, I want to sum all multiples of 3 and 5 under 1000. Here is my code:
int sum3n5(int max){
int sum = 0;
for(int i = 1; i <= max; ++i){
if( i%3 == 0 && i%5 == 0 ) { sum += i;}
else if( i%3 == 0 || i%5 == 0 ) { sum +=i;}
return sum;
};
};
Sorry if it is a trivial mistake that I failed to realize.
I always get the result 0 after running this.
int sum3n5(int max){
int sum = 0;
for (int i = 1; i <= max; ++i){
if( i % 3 == 0 || i % 5 == 0 ){
sum += i;
}
}
return sum;
}
You only need the || (logical or) operator, not the && (and certainly not both!). And the return needs to be after the for loop so that the loop can complete before the function returns.
A version without loop:
int sum3n5(int max)
{
return 3 * (max / 3) * (max / 3 + 1) / 2
+ 5 * (max / 5) * (max / 5 + 1) / 2
- 15 * (max / 15) * (max / 15 + 1) / 2;
}
It uses the fact that 1 + 2 + .. + n == n * (n + 1) / 2
How to create all possible numbers, starting from a given one, where all digits of the new ones are moved one slot to the right? For example if we have 1234. I want to generate 4123, 3412 and 2341.
What I have come out with so far is this:
int move_digits(int a)
{
int aux = 0;
aux = a % 10;
for(int i=pow(10, (number_digits(a) - 1)); i>0; i=i/10)
aux = aux * 10 + ((a % i) / (i/10));
return aux;
}
But it doesn't work.
The subprogram number_digits looks like this (it just counts how many digits the given number has):
int number_digits(int a)
{
int ct = 0;
while(a != 0)
{
a = a/10;
ct++;
}
return ct;
}
I think there is no need to write separate function number_digits.
I would write function move_digits simpler
#include <iostream>
#include <cmath>
int move_digits( int x )
{
int y = x;
double n = 0.0;
while ( y /= 10 ) ++n;
return ( x / 10 + x % 10 * std::pow( 10.0, n ) );
}
int main()
{
int x = 1234;
std::cout << x << std::endl;
std::cout << move_digits( x ) << std::endl;
}
Retrieving the last digit of n: n % 10.
To "cut off" the last digit, you could use number / 10.
Say you have a three-digit number n, then you can prepend a new digit d using 1000 * d + n
That said, you probably want to compute
aux = pow(10, number_digits - 1) * (aux % 10) + (aux / 10)
Calculatea/(number_digits(a) - 1) and a%(number_digits(a) - 1)
And your answer is (a%(number_digits(a) - 1))*10 + a/(number_digits(a) - 1)
int i =0 ;
int len = number_digits(a);
while(i < len){
cout << (a%(len - 1))*10 + a/(len - 1) <<endl;
a = (a%(len - 1))*10 + a/(len - 1);
}
void move_digits(int a)
{
int digits = 0;
int b = a;
while(b / 10 ){
digits++;
b = b / 10;
}
for (int i = 0; i < digits; ++i)
{
int c = a / 10;
int d = a % 10;
int res = c + pow(10, digits) * d;
printf("%d\n", res);
a = res;
}
printf("\n");
}
int main()
{
move_digits(12345);
}
How can I convert this recursive function to an iterative function?
#include <cmath>
int M(int H, int T){
if (H == 0) return T;
if (H + 1 >= T) return pow(2, T) - 1;
return M(H - 1, T - 1) + M(H, T - 1) + 1;
}
Well it's a 3-line code but it's very hard for me to convert this to an iterative function. Because it has 2 variables. And I don't know anything about Stacks so I couldn't convert that.
My purpose for doing this is speed of the function. This function is too slow. I wanted to use map to make this faster but I have 3 variables M, H and T so I couldn't use map
you could use dynamic programming - start from the bottom up when H == 0 and T == 0 calculate M and iterate them. here is a link explaining how to do this for Fibonacci numbers, which are quite similar to your problem.
Check this,recursive and not recursive versions gave equal results for all inputs i gave so far. The idea is to keep intermediate results in matrix, where H is row index, T is col index, and the value is M(H,T). By the way, you can calculate it once and later just obtain the result from the matrix, so you will have performance O(1)
int array[10][10]={{0}};
int MNR(int H, int T)
{
if(array[H][T])
return array[H][T];
for(int i =0; i<= H;++i)
{
for(int j = 0; j<= T;++j)
{
if(i == 0)
array[i][j] = j;
else if( i+1 > j)
array[i][j] = pow(2,j) -1;
else
array[i][j] = array[i-1][j-1] + array[i][j-1] + 1;
}
}
return array[H][T];
}
int M(int H, int T)
{
if (H == 0) return T;
if (H + 1 >= T) return pow(2, T) - 1;
return M(H - 1, T - 1) + M(H, T - 1) + 1;
}
int main()
{
printf("%d\n", M(6,3));
printf("%d\n", MNR(6,3));
}
Unless you know the formula for n-th (in your case, (m,n)-th) element of the sequence, the easiest way is to simulate the recursion using a stack.
The code should look like the following:
#include <cmath>
#include <stack>
struct Data
{
public:
Data(int newH, int newT)
: T(newT), H(newH)
{
}
int H;
int T;
};
int M(int H, int T)
{
std::stack<Data> st;
st.push(Data(H, T));
int sum = 0;
while (st.size() > 0)
{
Data top = st.top();
st.pop();
if (top.H == 0)
sum += top.T;
else if (top.H + 1 >= top.T)
sum += pow(2, top.T) - 1;
else
{
st.push(Data(top.H - 1, top.T - 1));
st.push(Data(top.H, top.T - 1));
sum += 1;
}
}
return sum;
}
The main reason why this function is slow is because it has exponential complexity, and it keeps recalculating the same members again and again. One possible cure is memoize pattern (handily explained with examples in C++ here). The idea is to store every result in a structure with a quick access (e.g. an array) and every time you need it again, retrieve already precomputed result. Of course, this approach is limited by the size of your memory, so it won't work for extremely big numbers...
In your case, we could do something like that (keeping the recursion but memoizing the results):
#include <cmath>
#include <map>
#include <utility>
std::map<std::pair<int,int>,int> MM;
int M(int H, int T){
std::pair<int,int> key = std::make_pair(H,T);
std::map<std::pair<int,int>,int>::iterator found = MM.find(key);
if (found!=MM.end()) return found->second; // skip the calculations if we can
int result = 0;
if (H == 0) result = T;
else if (H + 1 >= T) result = pow(2, T) - 1;
else result = M(H - 1, T - 1) + M(H, T - 1) + 1;
MM[key] = result;
return result;
}
Regarding time complexity, C++ maps are tree maps, so searching there is of the order of N*log(N) where N is the size of the map (number of results which have been already computed). There are also hash maps for C++ which are part of the STL but not part of the standard library, as was already mentioned on SO. Hash map promises constant search time (the value of the constant is not specified though :) ), so you might also give them a try.
You may calculate using one demintional array. Little theory,
Let F(a,b) == M(H,T)
1. F(0,b) = b
2. F(a,b) = 2^b - 1, when a+1 >= b
3. F(a,b) = F(a-1,b-1) + F(a,b-1) + 1
Let G(x,y) = F(y,x) ,then
1. G(x,0) = x // RULE (1)
2. G(x,y) = 2^x - 1, when y+1 >= x // RULE (2)
3. G(x,y) = G(x-1,y-1) + G(x-1,y) + 1 // RULE(3) --> this is useful,
// because for G(x,y) need only G(x-1,?), i.e if G - is two deminsions array, then
// for calculating G[x][?] need only previous row G[x-1][?],
// so we need only last two rows of array.
// Here some values of G(x,y)
4. G(0,y) = 2^0 - 1 = 0 from (2) rule.
5. G(1,0) = 1 from (1) rule.
6. G(1,y) = 2^1 - 1 = 1, when y > 0, from (2) rule.
G(0,0) = 0, G(0,1) = 0, G(0,2) = 0, G(0,3) = 0 ...
G(1,0) = 1, G(1,1) = 1, G(1,2) = 1, G(1,3) = 1 ...
7. G(2,0) = 2 from (1) rule
8. G(2,1) = 2^2 - 1 = 3 from (2) rule
9. G(2,y) = 2^2 - 1 = 3 when y > 0, from (2) rule.
G(2,0) = 2, G(2,1) = 3, G(2,2) = 3, G(2,3) = 3, ....
10. G(3,0) = 3 from (1) rule
11. G(3,1) = G(2,0) + G(2,1) + 1 = 2 + 3 + 1 = 6 from (3) rule
12. G(3,2) = 2^3 - 1 = 7, from (2) rule
Now, how to calculate this G(x,y)
int M(int H, int T ) { return G(T,H); }
int G(int x, int y)
{
const int MAX_Y = 100; // or something else
int arr[2][MAX_Y] = {0} ;
int icurr = 0, inext = 1;
for(int xi = 0; xi < x; ++xi)
{
for( int yi = 0; yi <= y ;++yi)
{
if ( yi == 0 )
arr[inext][yi] = xi; // rule (1);
else if ( yi + 1 >= xi )
arr[inext][yi] = (1 << xi) - 1; // rule ( 2 )
else arr[inext][yi] =
arr[icurr][yi-1] + arr[icurr][yi] + 1; // rule (3)
}
icurr ^= 1; inext ^= 1; //swap(i1,i2);
}
return arr[icurr][y];
}
// Or some optimizing
int G(int x, int y)
{
const int MAX_Y = 100;
int arr[2][MAX_Y] = {0};
int icurr = 0, inext = 1;
for(int ix = 0; ix < x; ++ix)
{
arr[inext][0] = ix; // rule (1)
for(int iy = 1; iy < ix - 1; ++ iy)
arr[inext][iy] = arr[icurr][iy-1] + arr[icurr][iy] + 1; // rule (3)
for(int iy = max(0,ix-1); iy <= y; ++iy)
arr[inext][iy] = (1 << ix ) - 1; // rule(2)
icurr ^= 1 ; inext ^= 1;
}
return arr[icurr][y];
}