I have to find all the prime numbers between two numbers m and n. (1 <= m <= n <= 1000000000 and n-m <= 100000). I am using sieve of eratosthenes but getting wrong answer. Can anyone help me what is wrong with my code.
#include<stdio.h>
#include<math.h>
int S[100002];
void sieve(long long int m, long long int n)
{
long long int x=sqrt(n);
long long int i,j;
long long int a;
for(i=0;i<=n-m+2;i++)
S[i]=0;
if(m%2==0)
i=m;
else {
i=m+1;
}
for (;i<=n;i+=2){
S[i-m]=1;
}
for (i=3;i<=x;i+=2){
if(i>=m && S[i-m]) continue;
if(i*i>=m)j=i*i;
else {
a = (m-i*i)%(2*i);
if(a==0)j=m;
else
j=m+ (2*i -a);
}
for (;j<=n;j+=2*i){
S[j-m]=1;
}
}
if (m==1)i=1; else i=0;
for (;i<=n-m;i++)
if (!S[i]){
printf("%lld\n",i+m);
}
}
int main(){
int t;
long long int m,n;
scanf("%d\n",&t);
while(t--){
scanf("%lld %lld",&m,&n);
sieve(m,n);
printf("\n");
}
return(0);
}
if(m%2==0)
i=m;
else {
i=m+1;
}
for (;i<=n;i+=2){
S[i-m]=1;
}
Now, what happens if m <= 2? Will 2 be considered prime or not?
You should use loop in main and call prime function.
For performance, I recommend you to avoid using sqrt function because it requires a lot of CPU clocks.
bool isPrime(int number){
if(number < 2) return false;
if(number == 2) return true;
if(number % 2 == 0) return false;
for(int i=3; (i*i)<=number; i+=2){
if(number % i == 0 ) return false;
}
return true;
}
***Change datatype for the range of number (long, long long, etc).
It is the most efficient(Sieve method) way of finding prime number between a range.
Here 1 is not consider as a prime number as a conventionally way.
#include<stdio.h>
#include<string.h>
#define max 10000000
using namespace std;
int main()
{
unsigned long long int i, j, k, m, n;
unsigned long long int* a = new unsigned long long int[max];
scanf("%ul %ul",&m,&n);
for(i = 1;i<=n;i++)
a[i]=i;
a[1] = 0;
for(i=2;(i*i)<=n;i++)
if(a[i]!=0)
for(k=2*i;k<=n;k=k+i)
if(a[k]!=0)
a[k]=0;
for(i =m;i<=n;i++)
if(a[i]!=0)
printf("%ul ",a[i]);
memset(a, 0, sizeof(a));
return 0;
}
#include<stdio.h>
#include<stdlib.h>
void prime(int );
int main(){
int x, end;
printf("Enter end of the range:\n");
scanf("%d", &end);
for(x = 2;x <= end;x++){
prime(x);
}
return 0;
}
void prime(int x){
int j, count = 1;
for(j=2;j <= x;j++){
if(x % j == 0){
count += 1;
//printf("count = %d,x = %d", count, x);
}
}
if(count == 2){
printf("\n%d\n", x);
}
}
Related
I have tried different inputs and when it exceeds a 7 or 8 digit value it just shows some wrong answers as outputs but it worked fine with most of my cases.
#include <stdio.h>
#include <stdlib.h>
int bin(unsigned long long int n){//gave function for binary convertion
if(n==0)
return 0;
else
return (n%2+10*bin(n/2));
}
int main()
{
unsigned long long int n,x;/*I even gave high digit data type*/
int i, v, count=0, max=0;
scanf("%llu",&n); /*if input is >8-digit output is wrong*/
x = bin(n);
v = floor(log10(x))+1; /*Its length*/
int a[v];
for(i = v-1; i >= 0; i--){ /*string it in array*/
a[i] = x%10;
x = x/10;
}
for(i = 0; i < v; i++){
if(a[i] == 0){
count = 0;}
else{
count++;}
if(max < count){
max = count;}
}
printf("%d",max);/*I gave 99999999 output is 8 but its shows 9*/
}
Your program has a number of problems. Here is one example:
int bin(unsigned long long int n){
^^^
The function returns an int so the calculation will overflow for even small numbers:
printf("%d\n", bin(1023)); // will print 1111111111 (fine)
printf("%d\n", bin(1024)); // will/may print 1410065408 (ups - very bad)
Even if you change to
unsigned long long int bin(unsigned long long int n){
overflow will happen soon.
In I'll recommend that you look directly into the binary pattern of the number using the & operator.
I'll not solve the complete task for you but here is some code that may help you.
#include <stdio.h>
#include <stdlib.h>
int main()
{
size_t t = 1;
size_t limit;
size_t n;
if (scanf("%zu", &n) != 1)
{
printf("Illegal input\n");
exit(1);
}
limit = 8 * sizeof n; // Assume 8 bit chars
for (size_t i = 0; i < limit; ++i)
{
if (n & t)
{
printf("Bit %zu is 1\n", i);
}
else
{
printf("Bit %zu is 0\n", i);
}
t = t << 1;
}
return 0;
}
What am i doing wrong here?
#include<stdio.h>
int main(){
int i,count;
long long int sum = 0, num;
for(num = 1; num <= 1000; num++){
count = 0;
for(i = 2; i <= num / 2; i++){
if(num % i == 0){
count++;
break;
}
}
if(count == 0 && num != 1)
sum = sum + num;
}
printf("Sum of prime numbers is: %d ", sum);
return 0;
}
I tried to make a program that outputs the sum of all primes below nth number, n being 2 million however when i try to run it, there is a slight delay and it outputs nothing... it works well enough for small numbers like 1000 or 100 but big numbers it just does not output anything.There are no error or bugs that i know of either. (please help, i know nothing, so guide this young one)
The above program is fine, but it is possible to further reduce the time complexity and make it faster. The better approach is to use sieve of eratosthenes algorithm.You can read about it here http://www.geeksforgeeks.org/sieve-of-eratosthenes/
The program for your problem is
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long int n)
{
bool prime[n+1];
memset(prime, true, sizeof(prime));
for (long long int p=2; p*p<=n; p++)
{
if (prime[p] == true)
{
for (int i=p*2; i<=n; i += p)
prime[i] = false;
}
}
long long sum=0;
for (int p=2; p<=n; p++)
if (prime[p])
sum+=p;
cout << sum << " ";
}
int main()
{
int n = 2000000;
cout << "sum of prime numbers less then 2000000 is :" << endl;
SieveOfEratosthenes(n);
return 0;
}
Please let me know if you face difficulty in understanding the program.Note: You can also use Segmented seive to further reduce the time complexity
#include<stdio.h>
#include<math.h>
int main(){
int i,count;
long long int sum=0, num;
for(num = 1;num<=1000000;num++){
count = 0;
for(i=2;i<=sqrt(num);i++){ //change here
if(num%i==0){
count++;
break;
}
}
if(count==0 && num!= 1)
sum = sum + num;
}
printf("Sum of prime numbers is: %lld ",sum); //and here
return 0;
}
Some simple maths, while checking for prime, check up to square root of a number. And in printf() the format specifier for long long int is %lld not %d.
I want to write a program to find all numbers less than an specific number that are equal to sum of a number and its polindrome.and this is my code, and it works correctly,but take too long time.what should I do to reduce it's time?
#include<stdio.h>
long long int isprime(long long int b)
{
long long int i,m;
for(i=2;i*i<=b;i++)
{
if(b%i==0)
{
return 0;
}
}
return 1;
}
long long int m(long long int a){
long long int l=0;
while (a>0){
l=l*10+(a%10);
a/=10;
}
return l;
}
int main(){
long long int i = 0,cnt = 0,a= 0,num;
scanf("%lld",&a);
for (i=2;i<=a; i++){
if (isprime(i) == 1){
for (num=1;num<i; num++){
if (num == m(i-num)){
cnt ++;
break;
//printf("%d\n",i);
}
}
}
}
printf("%lld", cnt);
}
How to compute factorial of numbers like 300 as the output is not even in bound of unsigned long long int ?Please help.
#include<stdio.h>
#include<stdlib.h>
unsigned long long int factorial(int number) {
unsigned long long int temp;
if(number <= 1) return 1;
temp = (number * factorial(number - 1));
return temp;
}
int main(){
int t,k,i,a[100001];
unsigned long long int sum[100001];
scanf("%d",&t);
for(i=0;i<t;i++){
scanf("%d",&a[i]);
}
for(k=0;k<t;k++){
sum[k]=0;
for(i=0;i<=a[k];i++){
sum[k] += ((factorial(a[k])/(factorial(i)*factorial(a[k]-i)))%3);
//printf("%d\n",sum[k]);
}}
for(i=0;i<t;i++){
printf("%llu\n",sum[i]);
}
return 0;
}
I tried this but it halt at only 60!.
You can try this for Factorials of large numbers :
#include<iostream>
#include<cstring>
int max = 5000;
void display(int arr[]){
int ctr = 0;
for (int i=0; i<max; i++){
if (!ctr && arr[i]) ctr = 1;
if(ctr)
std::cout<<arr[i];
}
}
void factorial(int arr[], int n){
if (!n) return;
int carry = 0;
for (int i=max-1; i>=0; --i){
arr[i] = (arr[i] * n) + carry;
carry = arr[i]/10;
arr[i] %= 10;
}
factorial(arr,n-1);
}
int main(){
int *arr = new int[max];
std::memset(arr,0,max*sizeof(int));
arr[max-1] = 1;
int num;
std::cout<<"Enter the number: ";
std::cin>>num;
std::cout<<"factorial of "<<num<<"is :\n";
factorial(arr,num);
display(arr);
delete[] arr;
return 0;
}
Factorial program using recursion in c with while loop.In this program once the execution reaches the function return statement it will not go back to the function call. Instead,it executes the function repeatedly. can anybody please tell me what's wrong in this program.
#include<stdio.h>
int fact(int n)
{
int x=1;
while(n>1)
{
x=n*fact(n-1);
}
return(x);
}
void main()
{
int n,fact1;
scanf("%d",&n);
fact1=fact(n);
printf("%d",fact1);
}
The reason that your program gets into an infinite loop is that the loop
while (n > 1)
x = n * fact(n-1);
never decrements n. Since n never decreases, the program will never leave the loop. Peter is correct in the comments: change the while to an if, and you will have a factorial function that handles all positive parameters correctly. However, even after changing while to if, your fact won't have the property that fact(0) == 1, as is required for a correct factorial function.
This
while(n>1)
is causing the looping. You don't change n inside the loop, so the loop is infinite.
Change while to if.
This is the method of factorial:
public int fact(int n)
{
if (n < 1)
{
return 1;
}
else
{
return n * fact(n - 1);
}
}
#include <stdio.h>
#include <stdlib.h>
/** main returns int, use it! */
int main(int argc, char **argv)
{
if (argc <= 2) {
if (argv) argc = atoi(argv[1] );
else return argc;
}
argc *= main (argc-1, NULL);
if (argv) {
printf("=%d\n", argc);
return 0;
}
return argc;
}
/*
Write a C++ Program to input a positive number,
Calculate and display factorial of this number
by recursion.
*/
#include<iostream.h>
#include<conio.h>
long factorial(int n);
void main()
{
clrscr();
int number, counter;
label1:
cout<<"\n Enter the Number = ";
cin>>number;
if ( number < 0)
{
cout<<"\n Enter a non negative number, please!";
goto label1;
}
cout<<"\n\n ----------- Results ------------";
cout<<"\n\n The Factorial of the number "<<number<<"\n is "<<factorial(number);
getch();
}
long factorial(int n)
{
if ( n == 0 )
return 1;
else
return n * factorial(n-1);
}
You can use the simple approach using recursion
#include <stdio.h>
int fact(int n)
{
if(n==1)
return 1;
else
return n * fact(n-1);
}
int main()
{
int f;
f = fact(5);
printf("Factorial = %d",f);
return 0;
}
Read it more C program to find factorial using recursion
/*several versions of a factorial program.*/
#include<stdio.h>
int main()
{
int n;
long factorial;
printf("Compute the factorial of what number? ");
scanf("%d", &n);
factorial = 1L;
while(n > 0)
factorial *= n--;
printf("The factorial is %ld\n", factorial);
return 0;
}
#include<stdio.h>
/*the same, but counting up to n instead of down to 0*/
int main()
{
register int count;
int n;
long factorial;
printf("Compute the factorial of what number? ");
scanf("%d", &n);
factorial = 1L;
count = 1;
while(count <= n)
factorial *= count++;
printf("%d! = %ld\n", n, factorial);
return 0;
}
#include<stdio.h>
/*an equivalent loop using 'for' instead of 'while'*/
int main()
{
register int count;
int n;
long factorial;
printf("Compute the factorial of what number? ");
scanf("%d", &n);
for(factorial = 1L, count = 1; count <= n; count++)
factorial *= count;
printf("%d! = %ld\n", n, factorial);
return 0;
}
/*WAP to find factorial using recursion*/
#include<stdio.h>
#include<stdlib.h>
int fact1=1;
int fact(int no)
{
fact1=fact1*no;
no--;
if(no!=1)
{
fact(no);
}
return fact1;
}
int main()
{
int no,ans;``
system("clear");
printf("Enter a no. : ");
scanf("%d",&no);
ans=fact(no);
printf("Fact : %d",ans);
return 0;
}
Factorial program using recursion in C with while loop.
int fact(int n)
{
int x=1;
while(n>=1)
{
return(n*fact(n-1));
}
return(1);
}
You can use this approach.
int factorial(int a)
{
while(a>1)
{
return a*factorial(a-1);
}
return 1;
}