Convert into binary and count the number of one’s . That is the answer
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#include<stdio.h>
int main()
{
short unsigned int t,i,n;
long unsigned int p;
scanf("%u",&t);
for(i=0;i<t;i++)
{
scanf("%ld",&p);
n = p/2048 + (p%2048)/1024 + ((p%2048)%1024)/512 + (((p%2048)%1024)%512)/256 + ((((p%2048)%1024)%512)%256)/128 + (((((p%2048)%1024)%512)%256)%128)/64 + ((((((p%2048)%1024)%512)%256)%128)%64)/32 + (((((((p%2048)%1024)%512)%256)%128)%64)%32)/16 + ((((((((p%2048)%1024)%512)%256)%128)%64)%32)%16)/8 + (((((((((p%2048)%1024)%512)%256)%128)%64)%32)%16)%8)/4 + ((((((((((p%2048)%1024)%512)%256)%128)%64)%32)%16)%8)%4)/2 + ((((((((((p%2048)%1024)%512)%256)%128)%64)%32)%16)%8)%4)%2;
printf("%u\n",n);
}
return 0;
}
runtime error… please help !!
1 Like
Write N as Q * 2048 + R, where Q and R are non-negative integers and R < 2048. Then the answer is Q + bitCount®, where bitCount(X) is the number of ones in binary representation of X.
This is my doubt as if N is less than 2048, we cant express the same as Q* 2048+ R as we are told that Q and R are integers( not a fraction ).
how to solve this problem in easy way please tell me…
can somebody please help me with my code here is the link : NovXIt - Online C++ Compiler & Debugging Tool - Ideone.com
i am getting wrong answer .
Why Isn’t THis Approach Working ??
#include <iostream>
#include <cmath>
using namespace std;
typedef long long int ll;
int main() {
ll t;
cin>>t;
while(t--)
{
ll ans=0,p,lpow;
cin>>p;
while(p>0)
{
lpow = log2(p);
p-=pow(2,lpow);
ans++;
}
cout<<ans<<endl;
}
return 0;
}Hi,
i have tried to solve using recursion
code follows,but not being accepted as solution
please help,thanks
int n,p,divi,rem,t;
int calc(int num){
if (num == 0)
return 0;
divi = num / 2048;
rem = num % 2048;
if (divi > 0)
t = divi + calc(rem);
else
t = calc(2 * rem);
return t;
}
int main(){
scanf("%d", &n);
for (int i = 0; i < n; i++){
scanf("%d", &p);
printf("%d \n", calc(p));
}
return 0;
}
#include<stdio.h>
int main()
{
int t,p,count,key;
scanf("%d",&t);
while(t–)
{
scanf("%d",&p);
if((p%2048)==0)
{
count=p/2048;
printf("%d",count);
printf("\n");
}
else
{
count=count_1§;
if(p/2048>1)
{
key=p/2048;
count=count+(key-1);
printf("%d",count);
printf("\n");
}
else
{
printf("%d",count);
printf("\n");
}
}
}
return 0;
}
int count_1(int n)
{
int ctd;
while(n)
{
n=n&(n-1);
ctd++;
}
return ctd;
}
it satisfy every case but during submission wrong answer
easiest way to solve the problem
import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner obj = new Scanner(System.in);
int a[] = new int[12];
for(int i=0;i<=11;i++){
a[i] = (int) Math.pow(2,i);
}
int n = obj.nextInt();
for(int i=1;i<=n;i++){
int p = obj.nextInt();
int k=0,l=0;
while(a[k]<=p){
l=k;
k++;
if(k==12){
break;}
}
int count=0;
while(a[l]>=1){
count= count+(p/a[l]);
p=p%a[l];
l=l-1;
if(l<0){
break;}
}
System.out.println(count);
}
}
}
i need to convert the set of number into “k”, “mil”, “bil” for my php website (Soundcloud downloader) when i implement this code the website automatically turns into error 500. Expert advice needed
I found one of the best solution for it, ie: convert decimal to binary and count the number of ones in it. that’s the answer.
for eg:
decimal: 253
binary: 11111101
no of ones: 7 that’s the answer.
For No > 2048, you have to use rest logic.
Thnx.
1 Like
Hey i tried the code using DP but the solution is not being accepted ,
the code works for the sample test cases and any other test case that i can think of .
Please help
Here is the link to my code :
https://www.codechef.com/viewsolution/14492082
can any one please check my code .I don’t know what is wrong it passed all sample tests.
https://www.codechef.com/viewsolution/17515413
Hello, guys, I have a website for downloading music from SoundCloud.
http://soundcloudmp3downloader.com/
And I hope u could help me… how can I get the information about bitrate of SC tracks?
is it possible? any idea?
My solution is a bit different and is very short if you are using C++ and stl as your tool to solve the problem
It uses a greedy approach to find the nearest value just smaller than or equal to n from the menu list.
while (n > 0) {
long nearest = upper_bound(menuList.begin(), menuList.end(), n) - menuList.begin();
n -= menuList[nearest - 1];
++count;
}
Here menuList is an array of the powers of 2 up to the required item - [1,2,4,8…]
int price[12] = {1, 2, 4, 8, 16, 32,
64, 128, 256, 512, 1024, 2048};
int t, p, c;
cin >> t;
for (int k = 0; k < t; k++)
{
c = 0;
cin >> p;
while(p != 0)
{
for (int i = 11; i >= 0; i--)
{
if (p >= price[i])
{
p -= price[i];
c++;
break;
}
}
}
cout << c << endl;
}
1 Like
if someone interested in dynamic programming solution
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
#define MAX 1000007
vector<int>v(12);
vector<int>dp(100002,-1);
int recu(int p)
{
if(p==0)
return 0;
if(p<0)
return MAX;
if(dp[p]!=-1)
return dp[p];
int q=MAX;
for(int i=0;i<12;i++)
{
q=min(q,recu(p-v[i])+1);
}
return dp[p]=q;
}
int main()
{
int t,p;
cin>>t;
dp[0]=0;
dp[1]=1;
for(int i=0;i<12;i++)
{
v[i]=(1<<i);
}
while(t--)
{
int ans=0;
cin>>p;
ans=recu(p);
cout<<ans<<"\n";
}
return 0;
}I am not familiar with Python but it seems that you are using file for input. You should read from the standard input and output to the standard output.
1 Like
I actually am using standard input. It reads from an input file on my computer (for testing), but if that file doesn’t exist (when I submit) it uses raw_input() instead. I’ve tested this method on the codechef judge and it works for other problems, so I don’t understand why it gets runtime error now.
1 Like