I attempted a Hackerearth tutorial question and was able to solve it correctly using following code:
#include <iostream>
using namespace std;
int main() {
const long int MOD = 1000000007;
const long int SIZE = 100001;
long int t, n;
long int cache[SIZE];
cache[0] = 1;
for (long int i = 1; i < SIZE; i++) {
cache[i] = ( i * cache[i-1] ) % MOD;
}
cin >> t;
while (t--) {
cin >> n;
cout << cache[n] << endl;
}
return 0;
}
However when using scientific notation to write MOD or SIZE, online judge reports incorrect answers. What am I missing here?
#include <iostream>
using namespace std;
int main() {
const long int MOD = 10e9+7;
const long int SIZE = 100001;
long int t, n;
long int cache[SIZE];
cache[0] = 1;
for (long int i = 1; i < SIZE; i++) {
cache[i] = ( i * cache[i-1] ) % MOD;
}
cin >> t;
while (t--) {
cin >> n;
cout << cache[n] << endl;
}
return 0;
}
Scientific notation means “constant multiplied by ten to the power of x”. If you want 1000000007 you need 1e9, meaning “one followed by nine zeroes.” Now you use 10e9 which is “ten followed by nine zeroes” or “one followed by ten zeroes” so you’re off by a factor of ten.
Related
I stuck at a problem SPOJ.
I checked all the test cases passing all of them, but I am still getting "WA" on spoj.
I know it can be solved using dynamic programming, but I am practicing memoization.
Here is my code:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector <int> dp(1000000);
long long int maxloot(vector<int> &loot, int n) {
if (n == 0)
return 0;
if (n == 1)
return loot[0];
if (n == 2)
return max(loot[0], loot[1]);
if (dp[n] != -1)
return dp[n];
long long int take = loot[n - 1] + maxloot(loot, n - 2);
long long int leave = maxloot(loot, n - 1);
return dp[n]= max(take, leave);
}
int main() {
int t;
cin >> t;
int p = 1;
while (t--) {
int n;
cin >> n;
vector <int> loot;
for (int i = 0; i < n; i++) {
int temp;
cin >> temp;
loot.push_back(temp);
}
dp.assign(1000000, -1);
cout <<"Case "<<p<<": "<< maxloot(loot, n)<<endl;
p++;
dp.clear();
}
}
Test case 1:
5
1 2 3 4 5
Test case 2:
1
10
output 1:
9
output 2:
10
You are using wrong data type to store value in vector dp.
As the sum of coins can go up to (10^9*10^2=10^11) therefore int would not be able to store it .Try using long long int instead as it would not lead to overflow condition.
SAME CODE AS YOURS(using long long int got accepted):
USE: vector< long long int>dp(1000000)
ACCEPTED CODE:
#include<iostream>
#include<vector>
#include<algorithm>
#define ull unsigned long long
using namespace std;
vector <long long int> dp(1000000);
long long int maxloot(vector<int> &loot, int n) {
if (n == 0)
return 0;
if (n == 1)
return loot[0];
if (n == 2)
return max(loot[0], loot[1]);
if (dp[n] != -1)
return dp[n];
long long int take = loot[n - 1] + maxloot(loot, n - 2);
long long int leave = maxloot(loot, n - 1);
return dp[n]= max(take, leave);
}
int main() {
int t;
cin >> t;
int p = 1;
while (t--) {
int n;
cin >> n;
vector <int> loot;
for (int i = 0; i < n; i++) {
int temp;
cin >> temp;
loot.push_back(temp);
}
dp.assign(1000000, -1);
cout <<"Case "<<p<<": "<< maxloot(loot, n)<<endl;
p++;
dp.clear();
}
}
I did implement Sieve prime generator. The code is pretty fast and consumes less memory. https://www.spoj.com/problems/PRIME1/
But I get "Wrong Answer" when I submit the solution. People online seem to just make the set the max to 32000 and run the solution. I don't exactly get where am I actually wrong? Or Is it just an extra newline (if possible) that makes the solution to be incorrect?
#include <iostream>
#include <unordered_map>
using namespace std;
int main(void) {
unordered_map<long long int, bool> notPrime;
notPrime[0] = true;
notPrime[1] = true;
for (long long int x = 2; x < 100000; x++) {
if (notPrime[x]) continue;
for (long long int u = 2 * x; u < 100000; u += x)
notPrime[u] = true;
}
int n;
cin >> n;
while (n--) {
long long int s, e;
cin >> s >> e;
if (s < 0)
s = 0;
for (long long int i = s; i <= e; i++) {
if (!notPrime[i]) {
cout << i << '\n';
}
}
if (n)
puts("");
}
return 0;
}
Screenshot of my code
Hey, I have just started learning C++ and I am trying to get it to sum the series:
K+N−1∑n=K [-1^(n)/(n+1)2]
I have managed to get it to tell me the nth term previously, but now I would like to for each term in the series, starting with the kth and going in sequence to the last (k+n-1st), add this term to the running sum.
I need to use a function direct_up() which uses the function term(). I defined initially and test it in the main.
I know I am missing something and am a bit confused about how to use the functions and that I may have made a few mistakes. So I would be very grateful for some advice or help. I have attached a picture of what I have done so far, as well as typed it below.
using namespace std;
double term(int n) {
double y;
if(n%2==1) {
y = -1.0/((n+1.0)*(n+1.0));
} else {
y = 1.0/((n+1.0)*(n+1.0));
}
return y;
}
double direct_up(int k, int n) {
int startingnumber = k;
for (int i = startingnumber; i <= k+n; i++) {
cout << n << term(n) << endl;
}
return n;
}
int main() {
double n;
int k;
cout.precision(16);
cout << "enter number of terms";
cin >> n;
cout << "enter a value for k";
cin >> k;
cout << "here is the value" << direct_up(k,n);
return 0;
}
This is what you want to do:
double direct_up(int k, int n)
{
int startingnumber = k;
double sum = 0;
for (int i = startingnumber; i <= k+n; i++) {
sum += term(i);
}
return sum;
}
Here's how to do it without keeping your own running sum as you asked in your comment:
#include <vector>
#include <numeric>
double direct_up(int k, int n)
{
int startingnumber = k;
std::vector<double> terms;
for (int i = startingnumber; i <= k+n; i++) {
terms.push_back(term(i));
}
return accumulate(terms.begin(), terms.end(), 0.0);
}
Here is the question:
Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 2^11=2048<3^7=2187.
However, confirming that 632382^518061>519432^525806 would be much more difficult, as both numbers contain over three million digits.
You are given N base exponent pairs, each forming a large number you have to find the Kth smallest number of them. K is 1−indexed.
Input Format
First line containts an integer N, number of base exponent pairs. Followed by N lines each have two space separated integers B and E, representing base and exponent.
Last line contains an integer K, where K<=N
Constraints
1≤N≤105
1≤K≤N
1≤B≤109
1≤E≤109
No two numbers are equal.
Here is my code:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,i = 0,k,j=0,x,m;
long long int *arr,*arr2,*arr3;
cin >> N;
arr = (long long int *)malloc(sizeof(long long int)*2*N);
arr2 = (long long int *)calloc(N,sizeof(long long int));
arr3 = (long long int *)calloc(N,sizeof(long long int));
x = 2*N;
while(x>0)
{
cin >> arr[i];
i++;
x--;
}
cin >> k;
for(i=0;i<2*N;i+=2)
{
arr2[j] = pow(arr[i],arr[i+1]);
j++;
}
arr3 = arr2;
sort(arr2,arr2+N);
for(i=0;i<N;i++)
{
if(arr3[i] == arr2[k-1])
{
m = i;
break;
}
}
cout << arr[2*m] << " " << arr[2*m + 1];
return 0;
}
The program works for small numbers only, can't make it work for large numbers. what to do?
Maybe you are generating an overflow on the big numbers. You could consider using a multiprecision arithmetic library such as https://gmplib.org/. I haven't used this library myself.
Have a look at this post How to detect integer overflow? on how to detect integer overflow.
From your choosing of long long int type I guess you calculated the a^b of the numbers in order to sort them, which leads to very big numbers and may lead to overflow.
Note that in order to sort the numbers there is no need for this calculation, for knowing if a^b > d^c it is sufficient to check log(a^b) > log(c^d) and therefore b*log(a) > d*log(c).
And it's better to use a struct or class to create a data structure for this big numbers.
This is the code for it:
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
struct BigNumber{
int base;
int exponent;
};
int Compare(BigNumber x, BigNumber y);
void Sort(BigNumber* arr, int N);
int main() {
int N,i = 0,k;
BigNumber *numbers;
cout<<"\nEnter N:";
cin >> N;
numbers = (BigNumber *)calloc(N,sizeof(BigNumber));
for(i=0; i<N; i++)
{
cout<<"\nEnter base and exponent for number "<<i<<":";
cin >> numbers[i].base>>numbers[i].exponent;
}
cout<<"\nEnter K:";
cin >> k;
Sort(numbers,N);
cout << "Kth number is :" << numbers[k].base << "^" << numbers[k].exponent;
return 0;
}
void Sort(BigNumber* arr, int N){
for(int i=0; i< N; i++ ){
for(int j=0; j< N; j++){
if(Compare(arr[i], arr[j])<0){
BigNumber temp = arr[j];
arr[j] = arr[i];
arr[i] = arr[j];
}
}
}
}
int Compare(BigNumber x, BigNumber y){
double X = x.exponent * log10(x.base);
double Y = y.exponent * log10(x.base);
return X == Y? 0: X > Y ? 1: -1;
}
I changed the code a little. Only problem I was having was I was calculating the exponent rather than comparing log of exponent.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,i = 0,k,j=0,x,m;
int *arr;
double *arr2,*arr3;
cin >> N;
arr = (int *)malloc(sizeof(int)*2*N);
arr2 = (double *)calloc(N,sizeof(double));
arr3 = (double *)calloc(N,sizeof(double));
x = 2*N;
while(x>0)
{
cin >> arr[i];
i++;
x--;
}
cin >> k;
for(i=0;i<2*N;i+=2)
{
arr2[j] = arr[i+1]*log10(arr[i]);
j++;
}
for (i = 0; i < N; i++) {
arr3[i] = arr2[i];
}
sort(arr2,arr2+N);
for(i=0;i<N;i++)
{
if(arr3[i] == arr2[k-1])
{
m = i;
break;
}
}
cout << arr[2*m] << " " << arr[2*m + 1];
return 0;
}
#include <iostream>
using namespace std;
unsigned long long divsum(unsigned long long x);
int main()
{
unsigned long long x;
cin >> x;
unsigned long long y[200000];
for (unsigned int i = 0; i < x; i++)
cin >> y[i];
for (unsigned int i = 0; i < x; i++){
cout << divsum(y[i]) << endl;
}
return 0;
}
unsigned long long divsum(unsigned long long x){
int sum = 0;
for(unsigned int i = 1; i <= x/2; i++){
if(x % i == 0)
sum += i;
}
return sum;
}
I'm doing an online exercise and it says there are possible 2000000 cases in the first line, so I made an array of that amount, however, when I submit the solution it exceeds the time.. so I was wondering what is an alternative and faster way to do this? The program works fine right now, except it exceeds the time limit of the website.
You can allocate the array dynamically, so it will work better in cases where x < 200000
int main()
{
unsigned long long x;
cin >> x;
unsigned long long *y = new unsigned long long[x];
// you can check if new didn't throw an exception here
for (unsigned int i = 0; i < x; i++)
cin >> y[i];
for (unsigned int i = 0; i < x; i++){
cout << divsum(y[i]) << endl;
}
delete[] y;
return 0;
}
Since you know the size of the array, try vector and reserve.
int main()
{
unsigned long long x;
cin >> x;
unsigned long long var;
vector<unsigned long long> y;
y.reserve(x);
for (unsigned int i = 0; i < x; i++){
cin >> y[i];
}for (unsigned int i = 0; i < x; i++){
cout << divsum(var) << endl;
}
return 0;
}
And deal in const &
const unsigned long long & divsum(const unsigned long long & x){
int sum = 0;
unsigned long long x2 = x/2
for(unsigned int i = 1; i <= x2; i++){
if(x % i == 0)
sum += i;
}
return sum;
}
I think your assignment is more complex than you think. Your divsum(x) function should return sum of all divisors of x, right? In this case it's better to factorize the x, and calculate this sum using all the prime numbers (with powers), product of which is equal to the x. Take a look at:
http://en.wikipedia.org/wiki/Divisor_function
There are methods for recursive factorization - for example, if you have factorized all numbers until n, you can quickly find factorization for (n + 1). You also need to generate primes, Erathosphene sieve for first 2000000 numbers would be fine.