This question already has answers here:
Why does pow(n,2) return 24 when n=5, with my compiler and OS?
(4 answers)
Closed 2 years ago.
For the below code:
int digsum(ll num) { //function to calculate sum of digits
if (num < 0)
num = abs(num);
int ans = 0;
while (num != 0) {
ans = ans + num % 10;
num = num / 10;
}
return ans;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
int a, b, c, cnt = 0;
long long x;
cin >> a >> b >> c;
for (int i = 0; i <= 72; i++) {
x = (b * (pow(i, a))) + c;
if (i == digsum(x) && x < mod) {
cout << x << " ";
}
}
return 0;
}
In the case a,b,c = 3,2,8 respectively and i=19;
pow(19,3) is supposed to calculate 19^3 but when I replace pow by (19x19x19), this specific case is getting satisfied, where as that wasn't the case with the pow function.
Can someone explain what the problem is?
My psychic powers suggest that your standard library's implementation of pow is not precise. I recall a discussion on SO a while back on this topic. Remember, pow returns a floating point value. I can't repro it, but it's entirely possibly your invocation of pow(19,3) returns 6858.999999999 or similar due to the way it's optimized.
Indeed, this this page says as much:
Due to rounding errors in floating point numbers, the results of pow() may not be precise (even if you pass it integers or whole numbers).
Also, this question and answer suggests the same thing.
I wouldn't have suspected it, but there you go.
Consider doing this as a workaround:
long long power = nearbyint(pow(i,a));
x = b * power + c;
Related
I am trying to solve a question in which i need to find out the number of possible ways to make a team of two members.(note: a team can have at most two person)
After making this code, It works properly but in some test cases it shows floating point error ad i can't find out what it is exactly.
Input: 1st line : Number of test cases
2nd line: number of total person
Thank you
#include<iostream>
using namespace std;
long C(long n, long r)
{
long f[n + 1];
f[0] = 1;
for (long i = 1; i <= n; i++)
{
f[i] = i * f[i - 1];
}
return f[n] / f[r] / f[n - r];
}
int main()
{
long n, r, m,t;
cin>>t;
while(t--)
{
cin>>n;
r=1;
cout<<C(n, min(r, n - r))+1<<endl;
}
return 0;
}
You aren't getting a floating point exception. You are getting a divide by zero exception. Because your code is attempting to divide by the number 0 (which can't be done on a computer).
When you invoke C(100, 1) the main loop that initializes the f array inside C increases exponentially. Eventually, two values are multiplied such that i * f[i-1] is zero due to overflow. That leads to all the subsequent f[i] values being initialized to zero. And then the division that follows the loop is a division by zero.
Although purists on these forums will say this is undefined, here's what's really happening on most 2's complement architectures. Or at least on my computer....
At i==21:
f[20] is already equal to 2432902008176640000
21 * 2432902008176640000 overflows for 64-bit signed, and will typically become -4249290049419214848 So at this point, your program is bugged and is now in undefined behavior.
At i==66
f[65] is equal to 0x8000000000000000. So 66 * f[65] gets calculated as zero for reasons that make sense to me, but should be understood as undefined behavior.
With f[66] assigned to 0, all subsequent assignments of f[i] become zero as well. After the main loop inside C is over, the f[n-r] is zero. Hence, divide by zero error.
Update
I went back and reverse engineered your problem. It seems like your C function is just trying to compute this expression:
N!
-------------
R! * (N-R)!
Which is the "number of unique sorted combinations"
In which case instead of computing the large factorial of N!, we can reduce that expression to this:
n
[ ∏ i ]
n-r
--------------------
R!
This won't eliminate overflow, but will allow your C function to be able to take on larger values of N and R to compute the number of combinations without error.
But we can also take advantage of simple reduction before trying to do a big long factorial expression
For example, let's say we were trying to compute C(15,5). Mathematically that is:
15!
--------
10! 5!
Or as we expressed above:
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15
-----------------------------------
1*2*3*4*5*6*7*8*9*10 * 1*2*3*4*5
The first 10 factors of the numerator and denominator cancel each other out:
11*12*13*14*15
-----------------------------------
1*2*3*4*5
But intuitively, you can see that "12" in the numerator is already evenly divisible by denominators 2 and 3. And that 15 in the numerator is evenly divisible by 5 in the denominator. So simple reduction can be applied:
11*2*13*14*3
-----------------------------------
1 * 4
There's even more room for greatest common divisor reduction, but this is a great start.
Let's start with a helper function that computes the product of all the values in a list.
long long multiply_vector(std::vector<int>& values)
{
long long result = 1;
for (long i : values)
{
result = result * i;
if (result < 0)
{
std::cout << "ERROR - multiply_range hit overflow" << std::endl;
return 0;
}
}
return result;
}
Not let's implement C as using the above function after doing the reduction operation
long long C(int n, int r)
{
if ((r >= n) || (n < 0) || (r < 0))
{
std::cout << "invalid parameters passed to C" << std::endl;
return 0;
}
// compute
// n!
// -------------
// r! * (n-r)!
//
// assume (r < n)
// Which maps to
// n
// [∏ i]
// n - r
// --------------------
// R!
int end = n;
int start = n - r + 1;
std::vector<int> numerators;
std::vector<int> denominators;
long long numerator = 1;
long long denominator = 1;
for (int i = start; i <= end; i++)
{
numerators.push_back(i);
}
for (int i = 2; i <= r; i++)
{
denominators.push_back(i);
}
size_t n_length = numerators.size();
size_t d_length = denominators.size();
for (size_t n = 0; n < n_length; n++)
{
int nval = numerators[n];
for (size_t d = 0; d < d_length; d++)
{
int dval = denominators[d];
if ((nval % dval) == 0)
{
denominators[d] = 1;
numerators[n] = nval / dval;
}
}
}
numerator = multiply_vector(numerators);
denominator = multiply_vector(denominators);
if ((numerator == 0) || (denominator == 0))
{
std::cout << "Giving up. Can't resolve overflow" << std::endl;
return 0;
}
long long result = numerator / denominator;
return result;
}
You are not using floating-point. And you seem to be using variable sized arrays, which is a C feature and possibly a C++ extension but not standard.
Anyway, you will get overflow and therefore undefined behaviour even for rather small values of n.
In practice the overflow will lead to array elements becoming zero for not much larger values of n.
Your code will then divide by zero and crash.
They also might have a test case like (1000000000, 999999999) which is trivial to solve, but not for your code which I bet will crash.
You don't specify what you mean by "floating point error" - I reckon you are referring to the fact that you are doing an integer division rather than a floating point one so that you will always get integers rather than floats.
int a, b;
a = 7;
b = 2;
std::cout << a / b << std::endl;
this will result in 3, not 3.5! If you want floating point result you should use floats instead like this:
float a, b;
a = 7;
b = 2;
std::cout << a / b << std::end;
So the solution to your problem would simply be to use float instead of long long int.
Note also that you are using variable sized arrays which won't work in C++ - why not use std::vector instead??
Array syntax as:
type name[size]
Note: size must a constant not a variable
Example #1:
int name[10];
Example #2:
const int asize = 10;
int name[asize];
I was given a task to write a program that displays:
I coded this:
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int a, n = 1, f = 1;
float s = 0;
cin >> a;
while(n <= a)
{
f = f * n;
s += 1 / (float)f;
n = n + 1;
}
cout << s;
getch();
}
So this displays -
s = 1 + 1/2! + 1/3! + 1/4! .... + 1/a!, including odd and even factorials.
For the past two hours I am trying to figure out how can I modify this code so that it displays the desired result. But I couldn't figure it out yet.
Question:
What changes should I make to my code?
You need to accumulate the sum while checking the counter n and only calculate the even factorials:
int n;
double sum = 1;
cin >> n;
for(int i = 2; i < n; ++i{
if(i % 2 == 0) sum += 1 / factorial(i);
}
In your code:
while(n <= a)
{
f = f * n;
// checks if n is even;
// n even if the remainder of the division by 2 is zero
if(n % 2 == 0){
s += 1 / (float)f;
}
n = n + 1;
}
12! is the largest value that fits in an 32 bit integer. You should use double for all the numbers. For even factorials, starting with f = 1 (0!), f = f * (n-1) * n, where n = 2, 4, 6, 8, ... .
You have almost everything you need in place (assuming you don't want to make design changes based on the issues brought up in the comments).
All you need to change is what you multiply f by in each step. To build up n! you are multiplying by n in each step. To build up (2n)! you would multiply by 2*n*(2*n-1)
Edit: Your second theory about what the instructor wants would need only slightly more of a change. Your inner loop could be replaced by
while(n < a)
{
f = f * n * (n+1);
s += 1 / f;
n = n + 2;
}
Edit2: To run your program I made several changes for I/O things you did that don't work in my copy of GCC. Hopefully those won't distract from the main point of the following code. I also added a second, more complicated and more accurate method of computing the answer to see how much was lost in floating point rounding.
So this code computes the answer twice, once by the method I suggested you change your code to and once by a more accurate method (using double instead of float and adding the numbers in the more accurate sequence via a recursive function). Then it display your answer and the difference between the two answers.
Running that shows the version I suggested gets all the displayed digits correct and is only wrong for the values of a I tried by tiny amounts that would need more display precision to notice:
#include<iostream>
using namespace std;
double fac_sum(int n, int a, double f)
{
if ( n > a )
return 0;
f *= n * (n-1);
return fac_sum(n+2, a, f) + 1 / f;
}
int main()
{
int a, n = 1;
float f = 1;
float s = 0;
cin >> a;
while(n < a)
{
f = f * n * (n+1);
s += 1 / f;
n = n + 2;
}
cout << s;
cout << " approx error was " << fac_sum( 2, a, 1.0)-s;
return 0;
}
For 8 that displays 0.54308 approx error was -3.23568e-08
I hope you understand the e-08 notation meaning the error is in the 8'th digit to the right of the .
Edit3: I changed f to float in this post because I had copied/tested thinking f was float, so parts of my answer didn't make sense when f was int
This question already has answers here:
How to calculate modulus of large numbers?
(10 answers)
Closed 9 years ago.
I have two numbers A and B
where A and B can be in the range 1<= A,B <=100^100000
How can we find the value of A^B modulo some M in C++ ??
In the duplicate I pointed out, the solution I particularly like is https://stackoverflow.com/a/8972838/1967396 (see there for attribution and references)
For your convenience I reproduce the code here (wrapped into an SCCE - but using C, not C++):
#include <stdio.h>
int modular(int base, unsigned int exp, unsigned int mod)
{
int x = 1;
int i;
int power = base % mod;
for (i = 0; i < sizeof(int) * 8; i++) {
int least_sig_bit = 0x00000001 & (exp >> i);
if (least_sig_bit)
x = (x * power) % mod;
power = (power * power) % mod;
}
return x;
}
int main(void) {
printf("123^456mod567 = %d\n", modular(123, 456, 567));
}
Amazing, isn't it.
use the formula (a*b)mod m = (a*(b (mod m))) (mod m). For more details see the wiki page Modular exponentiation
Another solution assumes that your M is fixed (or at least that you need to compute A^B many times with the same M).
Step 1: compute the Euler's totient function (this requires a factorization of M, so it's quite expensive). Let's call this number k.
Due to the Fermat's little theorem, your answer is simply:
(a % M)^(b % k)
Now, unless M is a large prime number, this greatly simplify the problem.
The above problem can be solved using the code snippet below.
Thus to ensure this code does not overflow, check that n * n <= 2 ^ 31.
int modPow(int base, int exp, int n) {
base = base%n;
if (exp == 0)
return 1;
else if (exp == 1)
return base;
else if (exp%2 == 0)
return modPow(base*base%n, exp/2, n);
else
return base*modPow(base, exp-1, n)%n;
}
I am doing a factorial program with strings because i need the factorial of Numbers greater than 250
I intent with:
string factorial(int n){
string fact="1";
for(int i=2; i<=n; i++){
b=atoi(fact)*n;
}
}
But the problem is that atoi not works. How can i convert my string in a integer.
And The most important Do I want to know if the program of this way will work with the factorial of 400 for example?
Not sure why you are trying to use string. Probably to save some space by not using integer vector? This is my solution by using integer vector to store factorial and print.Works well with 400 or any large number for that matter!
//Factorial of a big number
#include<iostream>
#include<vector>
using namespace std;
int main(){
int num;
cout<<"Enter the number :";
cin>>num;
vector<int> res;
res.push_back(1);
int carry=0;
for(int i=2;i<=num;i++){
for(int j=0;j<res.size();j++){
int tmp=res[j]*i;
res[j]=(tmp+carry)%10 ;
carry=(tmp+carry)/10;
}
while(carry!=0){
res.push_back(carry%10);
carry=carry/10;
}
}
for(int i=res.size()-1;i>=0;i--) cout<<res[i];
cout<<endl;
return 0;
}
Enter the number :400
Factorial of 400 :64034522846623895262347970319503005850702583026002959458684445942802397169186831436278478647463264676294350575035856810848298162883517435228961988646802997937341654150838162426461942352307046244325015114448670890662773914918117331955996440709549671345290477020322434911210797593280795101545372667251627877890009349763765710326350331533965349868386831339352024373788157786791506311858702618270169819740062983025308591298346162272304558339520759611505302236086810433297255194852674432232438669948422404232599805551610635942376961399231917134063858996537970147827206606320217379472010321356624613809077942304597360699567595836096158715129913822286578579549361617654480453222007825818400848436415591229454275384803558374518022675900061399560145595206127211192918105032491008000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
There's a web site that will calculate factorials for you: http://www.nitrxgen.net/factorialcalc.php. It reports:
The resulting factorial of 250! is 493 digits long.
The result also contains 62 trailing zeroes (which constitutes to 12.58% of the whole number)
3232856260909107732320814552024368470994843717673780666747942427112823747555111209488817915371028199450928507353189432926730931712808990822791030279071281921676527240189264733218041186261006832925365133678939089569935713530175040513178760077247933065402339006164825552248819436572586057399222641254832982204849137721776650641276858807153128978777672951913990844377478702589172973255150283241787320658188482062478582659808848825548800000000000000000000000000000000000000000000000000000000000000
Many systems using C++ double only work up to 1E+308 or thereabouts; the value of 250! is too large to store in such numbers.
Consequently, you'll need to use some sort of multi-precision arithmetic library, either of your own devising using C++ string values, or using some other widely-used multi-precision library (GNU GMP for example).
The code below uses unsigned double long to calculate very large digits.
#include<iostream.h>
int main()
{
long k=1;
while(k!=0)
{
cout<<"\nLarge Factorial Calculator\n\n";
cout<<"Enter a number be calculated:";
cin>>k;
if (k<=33)
{
unsigned double long fact=1;
fact=1;
for(int b=k;b>=1;b--)
{
fact=fact*b;
}
cout<<"\nThe factorial of "<<k<<" is "<<fact<<"\n";
}
else
{
int numArr[10000];
int total,rem=0,count;
register int i;
//int i;
for(i=0;i<10000;i++)
numArr[i]=0;
numArr[10000]=1;
for(count=2;count<=k;count++)
{
while(i>0)
{
total=numArr[i]*count+rem;
rem=0;
if(total>9)
{
numArr[i]=total%10;
rem=total/10;
}
else
{
numArr[i]=total;
}
i--;
}
rem=0;
total=0;
i=10000;
}
cout<<"The factorial of "<<k<<" is \n\n";
for(i=0;i<10000;i++)
{
if(numArr[i]!=0 || count==1)
{
cout<<numArr[i];
count=1;
}
}
cout<<endl;
}
cout<<"\n\n";
}//while
return 0;
}
Output:
![Large Factorial Calculator
Enter a number be calculated:250
The factorial of 250 is
32328562609091077323208145520243684709948437176737806667479424271128237475551112
09488817915371028199450928507353189432926730931712808990822791030279071281921676
52724018926473321804118626100683292536513367893908956993571353017504051317876007
72479330654023390061648255522488194365725860573992226412548329822048491377217766
50641276858807153128978777672951913990844377478702589172973255150283241787320658
18848206247858265980884882554880000000000000000000000000000000000000000000000000
000000000000][1]
You can make atoi compile by adding c_str(), but it will be a long way to go till getting factorial. Currently you have no b around. And if you had, you still multiply int by int. So even if you eventually convert that to string before return, your range is still limited. Until you start to actually do multiplication with ASCII or use a bignum library there's no point to have string around.
Your factorial depends on conversion to int, which will overflow pretty fast, so you want be able to compute large factorials that way. To properly implement computation on big numbers you need to implement logic as for computation on paper, rules that you were tought in primary school, but treat long long ints as "atoms", not individual digits. And don't do it on strings, it would be painfully slow and full of nasty conversions
If you are going to solve factorial for numbers larger than around 12, you need a different approach than using atoi, since that just gives you a 32-bit integer, and no matter what you do, you are not going to get more than 2 billion (give or take) out of that. Even if you double the size of the number, you'll only get to about 20 or 21.
It's not that hard (relatively speaking) to write a string multiplication routine that takes a small(ish) number and multiplies each digit and ripples the results through to the the number (start from the back of the number, and fill it up).
Here's my obfuscated code - it is intentionally written such that you can't just take it and hand in as school homework, but it appears to work (matches the number in Jonathan Leffler's answer), and works up to (at least) 20000! [subject to enough memory].
std::string operator*(const std::string &s, int x)
{
int l = (int)s.length();
std::string r;
r.resize(l);
std::fill(r.begin(), r.end(), '0');
int b = 0;
int e = ~b;
const int c = 10;
for(int i = l+e; i != e;)
{
int d = (s[i]-0x30) * x, p = i + b;
while (d && p > e)
{
int t = r[p] - 0x30 + (d % c);
r[p] = (t % c) + 0x30;
d = t / c + d / c;
p--;
}
while (d)
{
r = static_cast<char>((d % c) +0x30)+r;
d /= c;
b++;
}
i--;
}
return r;
}
In C++, the largest integer type is 'long long', and it hold 64 bits of memory, so obviously you can't store 250! in an integer type. It is a clever idea to use strings, but what you are basically doing with your code is (I have never used the atoi() function, so I don't know if it even works with strings larger than 1 character, but it doesn't matter):
covert the string to integer (a string that if this code worked well, in one moment contains the value of 249!)
multiply the value of the string
So, after you are done multiplying, you don't even convert the integer back to string. And even if you did that, at one moment when you convert the string back to an integer, your program will crash, because the integer won't be able to hold the value of the string.
My suggestion is, to use some class for big integers. Unfortunately, there isn't one available in C++, so you'll have to code it by yourself or find one on the internet. But, don't worry, even if you code it by yourself, if you think a little, you'll see it's not that hard. You can even use your idea with the strings, which, even tough is not the best approach, for this problem, will still yield the results in the desired time not using too much memory.
This is a typical high precision problem.
You can use an array of unsigned long long instead of string.
like this:
struct node
{
unsigned long long digit[100000];
}
It should be faster than string.
But You still can use string unless you are urgent.
It may take you a few days to calculate 10000!.
I like use string because it is easy to write.
#include <bits/stdc++.h>
#pragma GCC optimize (2)
using namespace std;
const int MAXN = 90;
int n, m;
int a[MAXN];
string base[MAXN], f[MAXN][MAXN];
string sum, ans;
template <typename _T>
void Swap(_T &a, _T &b)
{
_T temp;
temp = a;
a = b;
b = temp;
}
string operator + (string s1, string s2)
{
string ret;
int digit, up = 0;
int len1 = s1.length(), len2 = s2.length();
if (len1 < len2) Swap(s1, s2), Swap(len1, len2);
while(len2 < len1) s2 = '0' + s2, len2++;
for (int i = len1 - 1; i >= 0; i--)
{
digit = s1[i] + s2[i] - '0' - '0' + up; up = 0;
if (digit >= 10) up = digit / 10, digit %= 10;
ret = char(digit + '0') + ret;
}
if (up) ret = char(up + '0') + ret;
return ret;
}
string operator * (string str, int p)
{
string ret = "0", f; int digit, mul;
int len = str.length();
for (int i = len - 1; i >= 0; i--)
{
f = "";
digit = str[i] - '0';
mul = p * digit;
while(mul)
{
digit = mul % 10 , mul /= 10;
f = char(digit + '0') + f;
}
for (int j = 1; j < len - i; j++) f = f + '0';
ret = ret + f;
}
return ret;
}
int main()
{
freopen("factorial.out", "w", stdout);
string ans = "1";
for (int i = 1; i <= 5000; i++)
{
ans = ans * i;
cout << i << "! = " << ans << endl;
}
return 0;
}
Actually, I know where the problem raised At the point where we multiply , there is the actual problem ,when numbers get multiplied and get bigger and bigger.
this code is tested and is giving the correct result.
#include <bits/stdc++.h>
using namespace std;
#define mod 72057594037927936 // 2^56 (17 digits)
// #define mod 18446744073709551616 // 2^64 (20 digits) Not supported
long long int prod_uint64(long long int x, long long int y)
{
return x * y % mod;
}
int main()
{
long long int n=14, s = 1;
while (n != 1)
{
s = prod_uint64(s , n) ;
n--;
}
}
Expexted output for 14! = 87178291200
The logic should be:
unsigned int factorial(int n)
{
unsigned int b=1;
for(int i=2; i<=n; i++){
b=b*n;
}
return b;
}
However b may get overflowed. So you may use a bigger integral type.
Or you can use float type which is inaccurate but can hold much bigger numbers.
But it seems none of the built-in types are big enough.
I wrote the following code to sum the series (-1)^i*(i/(i+1)). But when I run it I get -1 for any value of n.
Can some one please point out what I am doing wrong? Thank you in advance!
#include <iostream>
using namespace std;
int main()
{
int sum = 0;
int i = 1.0;
int n = 5.0;
for(i=1;i<=n;i++)
sum = (-1)^i*(i/(i+1));
cout << "Sum" <<" = "<< sum << endl;
return 0;
}
Problem #1: The C++ ^ operator isn't the math power operator. It's a bitwise XOR.
You should use pow() instead.
Problem #2:
You are storing floating-point types into an integer type. So the following will result in integer division (truncated division):
i/(i+1)
Problem #3:
You are not actually summing anything up:
sum = ...
should be:
sum += ...
A corrected version of the code is as follows:
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++)
sum += pow(-1.,(double)i) * ((double)i / (i + 1));
Although you really don't need to use pow in this case. A simple test for odd/even will do.
double sum = 0;
int i = 1;
int n = 5;
for(i = 1; i <= n; i++){
double val = (double)i / (i + 1);
if (i % 2 != 0){
val *= -1.;
}
sum += val;
}
You need too put sum += pow(-1,i)*(i/(i+1));
Otherwise you lose previous result each time.
Use pow function for pow operation.
edit : as said in other post, use double or float instead of int to avoid truncated division.
How about this
((i % 2) == 0 ? 1 : -1)
instead of
std::pow(-1, i)
?
Full answer:
double sum = 0;
int i = 1.0;
int n = 5.0;
for (i = 1; i <= n; ++i) {
signed char sign = ((i % 2) == 0 ? 1 : -1);
sum += sign * (i / (i+1));
}
Few problems:
^ is teh bitwise exclusive or in c++ not "raised to power". Use pow() method.
Remove the dangling opening bracket from the last line
Use ints not floats when assigning to ints.
You seem to have a few things wrong with your code:
using namespace std;
This is not directly related to your problem at hand, but don't ever say using namespace std; It introduces subtle bugs.
int i = 1.0;
int n = 5.0;
You are initializaing integral variables with floating-point constants. Try
int i = 1;
int n = 5;
sum = (-1)^i*(i/(i+1));
You have two problems with this expression. First, the quantity (i/(i+1)) is always zero. Remember dividing two ints rounds the result. Second, ^ doesn't do what you think it does. It is the exclusive-or operator, not the exponentiation operator. Third, ^ binds less tightly than *, so your expression is:
-1 xor (i * (i/(i+1)))
-1 xor (i * 0)
-1 xor 0
-1
^ does not do what you think it does. Also there are some other mistakes in your code.
What it should be:
#include <iostream>
#include <cmath>
int main( )
{
long sum = 0;
int i = 1;
int n = 5;
for( i = 1; i <= n; i++ )
sum += std::pow( -1.f, i ) * ( i / ( i + 1 ) );
std::cout << "Sum = " << sum << std::endl;
return 0;
}
To take a power of a value, use std::pow (see here). Also you can not assign int to a decimal value. For that you need to use float or double.
The aforementioned ^ is a bitwise-XOR, not a mark for an exponent.
Also be careful of Integer Arithmetic as you may get unexpected results. You most likely want to change your variables to either float or double.
There are a few issues with the code:
int sum = 0;
The intermediate results are not integers, this should be a double
int i = 1.0;
Since you will use this in a division, it should be a double, 1/2 is 0 if calculated in integers.
int n = 5.0;
This is an int, not a floating point value, no .0 is needed.
for(i=1;i<=n;i++)
You've already initialized i to 1, why do it again?
sum = (-1)^i*(i/(i+1));
Every iteration you lose the previous value, you should use sum+= 'new values'
Also, you don't need pow to calculate (-1)^i, all this does is switch between +1 and -1 depending on the odd/even status of i. You can do this easier with an if statement or with 2 for's, one for odd i one for even ones... Many choices really.