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name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | # AOJ 0053: Sum of Prime Numbers
# Python3 2018.6.15 bal4u
MAX = 104729 # 10000th prime
SQRT = 323 # sqrt(MAX)
prime = [True for i in range(MAX+2)]
def sieve():
for i in range(2, MAX, 2):
prime[i] = False
for i in range(3, SQRT, 2):
if prime[i] is True:
for j in range(i*i, MAX, i):
prime[j] = False
... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <cmath>
#include <iostream>
#include <vector>
typedef std::vector <int> vi;
vi P;
vi f(int n){
int i=0,j=1,h=(n+1)/2,x,r=(int)pow(n,.5);
vi s(h),p(1);
while(j<=n){
s[i++]=j;
j+=2;
}
p[0]=2;
for (i=1;i<h;i++)
if (x=s[i]){
p.push_back(x);
if (i<=r) for (j=x*x/2;j<h;j+=x) s[j]=0;
}
return p;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long lli;
const int MAXP = 200000;
bool isP[MAXP];
vector<lli> Prime;
int main() {
fill(isP, isP+MAXP, true);
isP[0] = isP[1] = false;
for(int i = 2; i < MAXP; ++i) {
if(!isP[i]) continue;
Prime.push_back(i);
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | # Aizu Problem 0053: Sum of Prime Numbers
#
import sys, math, os, bisect
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def primes2(n):
""" Input n>=6, Returns a list of primes, 2 <= p < n """
n, correction = n-n%6+6, 2-(n%6>1)
sieve = [True... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.*;
import java.io.*;
public class Main {
public static boolean[] sieve (int n) {
boolean[] isPrime = new boolean[n];
Arrays.fill(isPrime, true);
int lim = (int)Math.sqrt(n);
for (int i = 4; i < isPrime.length; i += 2)
isPrime[i] = false;
for (int i = 3; i <= lim; i += 2) {
... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <cmath>
using namespace std;
int main(){
int n, prime[10000], cnt = 1;
prime[0] = 2;
for (int i = 3; cnt < 10000; i += 2){
int k = 0;
for (int j = 3; j <= sqrt(i); j++) {
if (i % j == 0){
k = 1;
break;
}
}
if (!k) {
prime[cnt] = prime[cnt - 1] + i;
cnt++;
}... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main (){
int arr[1000000];
int N=1000000;
for(int i = 0; i < N; i++){
arr[i] = 1;
}
for(int i = 2; i < sqrt(N); i++){
if(arr[i]){
for(int j = 0; i * (j + 2) < N; j++){
arr[i *(j + 2)] = 0;
}
}
}
while(1... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.ArrayList;
import java.util.Scanner;
//import java.io.*;
//import java.util.Arrays;
public class Main {
public static void main(String[] args) throws java.io.IOException {
Scanner scan = new Scanner(System.in);
//InputStreamReader is = new InputStreamReader(System.in);
//BufferedReader br = new ... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys
def prime(m):
N=range(1,m+2,2)
r=int(m**.5)
h=len(N)
N[0]=0
for i in range(h):
x=N[i]
if x>r: break
if x and i+x<h: N[i+x:h:x]=[0] * ((h-1-i-x)/x+1)
N[0]=2
return filter(None,N)
PRIMES=prime(105000)
for n in sys.stdin:
n=int(n)
if n==0: break
pr... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
bool ar[110000]={true};
int prime(int x){
int sump=0;
int i=110000;
for(int a=0;a<i;a++){ar[a]=true;}
for(int b=2;b*b<=i;b++){
if(ar[b]){
for(int c=b*b;c<=i;c+=b){
ar[c]=false;
}
}
}
int ca=0;
for(int d=2;d<=i;d++){
if(ar[d]){
sump+=d;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Arrays;
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
boolean index[]=new boolean[1000000];
Arrays.fill(index, true);
index[0]=false;
index[1]=false;
for(int i=0;i<index.length;i++){
if(index[i]){
for(int j... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
static boolean[] isPrime = new boolean[1000000];
static int[] a;
static int[] sum;
static{
Arrays.fill(isPrime, true);
isPrime[0]=isPrime[1]=false;
int count=0;
int[] b = new int[100000];
for(int i = 2; i*... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import math
prime = [0 for i in range(11000)]
k = 0
r = 105000
sqrt = int(math.sqrt(r))
p = [1]*r
p[0] = 0
for i in range(1,sqrt):
if p[i]:
prime[k] = i+1
k += 1
l = i
for j in range(2*i+1,r,i+1):
p[j] = 0
for i in range(l+1,r):
if p[i]:
prime[k] = i+1
... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class Main
{
private static ArrayList<Integer> primeNumberList = new ArrayList<>();
public static void main(String[] args) throws NumberFormatException, IOException
{
BufferedReade... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<cstdio>
int p[1<<17],s,i,n;
int main(){for(i=2;i<1<<17;i++)for(s=2;!p[i]&&i*s<1<<17;p[i*s++]++);for(;scanf("%d",&n),n;printf("%d\n",s))for(i=2,s=0;(!p[i])?s+=i,n--:n,n;i++);} | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | def prime(m):
N=range(1,m+2,2)
r=int(m**.5)
h=len(N)
N[0]=0
for i in range(h):
x=N[i]
if x>r: break
if x and i+x<h: N[i+x:h:x]=[0] * ((h-1-i-x)/x+1)
N[0]=2
return filter(None,N)
PRIMES=prime(105000)
while 1:
n=input()
if n==0: break
print sum(PRIMES[:n]) | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
bool isprime[1000001];
int main(){
vector<int> v;
// init
for(int i = 0 ; i <= 1000000 ; i++)
isprime[i] = true;
isprime[0] = isprime[1] = false;
// eratosuteneru
for(int i = 2 ; i*i <= 1000000 ; i++){
if( isprime[i] ){
for(int j = i*i ; j <= ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<cstdio>
#include<iostream>
#include<string>
#define NMAX (10000+10)
#define YES 0
#define NO 1
using namespace std;
int main(void){
int huriko[NMAX*20]={YES};
int cnt=0;
long long primenum[NMAX];
huriko[0]=NO,huriko[1]=NO;
primenum[0]=0;
for(int i=2;cnt<NMAX-5;i++){
if(huriko[i]==YES){
cnt++;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | def primes(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
if __name... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | is_prime = [1] * 200000
primes = []
for i in range(2,200000):
if is_prime [i] == False:
continue
primes.append(i)
for j in range(i * 2,200000,i):
is_prime [j] = False
while True:
N = int(input())
if N == 0:
break
print(sum(primes [:N])) | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Scanner;
public class Main {
public static final int MAX = 200000;
public static boolean[] isPrime = new boolean[MAX + 1];
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
eratos();
int n = sc.nextInt();
while (n != 0) {
int sum = 0, count = 0;
for... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | def sieve(n):
num = [True] * n
num[0] = num[1] = False
for i in range(2,int(n**0.5)+1):
if num[i]:
for j in range(i**2, n, i):
num[j] = False
return [i for i in range(2, n) if num[i]]
prime = sieve(110000)
while True:
n = int(input())
if n == 0: break
print(sum(p... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
using namespace std;
int main(){
int prime[10001],p = 3,psum[10001],n;
prime[1] = 2,prime[2] = 3;
for(int i=5;p < 10001;i++){
int f = 0;
for(int j=1;prime[j]*prime[j] <= i;j++) if(i % prime[j] == 0) {f = 1;break;}
if(!f) prime[p++] = i;
}
psum[1] = 2;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> prime;
bool isPrime[1000001];
void erats(){
fill(isPrime,isPrime+1000001,true);
isPrime[0]=false;
isPrime[1]=false;
for(int i = 2; i < 1000000; i++){
if(isPrime[i]){
prime.push_back(i);
for(int j = i*2; j < 100000... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.*;
import java.util.stream.*;
import java.io.*;
import java.math.*;
public class Main {
static void solve (FastScanner in, PrintWriter out, Methods ms) {
ArrayList<Integer> primeList = getPrimeList(ms);
while (true) {
int n = in.nextInt();
if (n == 0) return;
long sum = 0;
... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.io.*;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.NoSuchElementException;
import java.util.Scanner;
public class Main {
static ArrayList<Integer> listX = new ArrayList<Integer>();
public stati... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#define N 110000
using namespace std;
int main(){
bool num[N] ;
for(int i =0;i<N;i++)
num[i] = true;
num[0] = false; num[1] = false;
for(int i=2;i*i<N;i++){
if(num[i] == true){
for(int k = i * 2;k<N;k +=i)
num[k] = false;
}
}
int n;
while(1){
cin >> n;
if(n == 0){
return 0;
}
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
def make_prime_list(lim):
lst = [True for i in xrange(lim + 1)]
prime_list = []
num = 2
while num ** 2 <= lim:
for i in xrange(num * 2, lim + 1, num):
lst[i] = False
num += 1
while not lst[num]:
nu... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
from __future__ import (division, absolute_import, print_function,
unicode_literals)
from sys import stdin
def enum_prime(limit):
L = list(range(2, limit + 1))
P = []
while L[0] ** 2 < limit:
P.append(L[0])
L = [i for i i... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
using namespace std;
int isprime(int n)
{
int i=3;
while(i*i<=n){
if(n%i==0) return false;
i+=2;
}
return true;
}
int ans(int n)
{
int i=1;
int a=2;
int k=3;
while... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<cstdio>
#include<vector>
using namespace std;
int n;
vector<int> p;
int sosu[500000];
int main(){
sosu[1]=1;
for(int i=2;i<400000;i++){
if(sosu[i]==1) continue;
for(int j=i*2;j<400000;j+=i)
sosu[j]=1;
}
for(int i=2;i<400000;i++){
if(!sosu[i]) p.push_back(i);
}
while(1){
scanf... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | isPrime = [True] * 200001
sum = 0
sumPrime = [sum]
def calcPrime():
global sum
isPrime[0] = isPrime[1] = False
for i in range(len(isPrime)):
if(isPrime[i]):
sum += i
sumPrime.append(sum)
for j in range(2 * i, len(isPrime), i):
isPrime[j] = False
... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | from itertools import *
n=range(104730);a=list(n)
for i in range(2,321):a[i*2::i]=[0]*len(a[i*2::i])
p=list(compress(n,a))
for e in iter(input,'0'):print(sum(p[:int(e)+1])-1)
| PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<cstdio>
#include<algorithm>
using namespace std;
bool p[500000];
int s[100000];
void set()
{
fill(p,p+500000,true);
p[0]=false;
p[1]=false;
for(int i=2;i<500000;i++)
{
for(int j=2;j*i<500000;j++)
{
p[i*j]=false;
}
}
int count=0;
for(int i=0;i<500000;i++)
{
if(p[i])
{
s[count]=i;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
using namespace std;
bool p[1000000];
int main()
{
for (int i = 2; i < 1000000; i++){
p[i] = true;
}
for (int i = 2; i * i < 1000000; i++){
if (p[i]){
for (int j = i * 2; j < 1000000; j += i){
p[j] = false;
}
}
}
int n;
while (cin >> n, n){
int c = 0;
long long sum ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | from itertools import *
n=104730;a=list(range(n));a[:2]=0,0
for i in range(2,323):a[i*2::i]=[0]*len(a[i*2::i])
p=list(compress(range(n),a))
for e in iter(input,'0'):print(sum(p[:int(e)]))
| PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.io.*;
class Main {
private static final int M = 104730;
public static void main (String args[]) {
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(isr);
String buf;
int list[] = new int[M+1];
for (int i=2;i<=M;i++) list[i] = 0;
for (int i=2;i<... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
bool isPrime(int n){
if(n<=1 || n!=2&&(n&1)==0)return false;
for(int i=3;i*i<=n;i+=2)
if(n%i==0)return false;
return true;
}
int main(){
int a[10001];
int n,m=1;
a[0]=0;a[1]=2;
for(int i=3;m<10000;i+=2){
if(isPrime(i)){
a[m+1]=i+a[m];
m++;
}
}
while(cin>>n&&n... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX 1000000
int main(){
bool p[MAX+1];
int n, sum;
p[0] = false;
p[1] = false;
for(int i=2 ; i<=MAX ; i++){
p[i] = true;
}
for(int i=2 ; i<=sqrt(MAX)+1 ; i++){
if(p[i]==1){
for(int j=i*2; j<=MAX; j=j+i){
p[j] = false;
}
}
}
w... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class Main
{
private static ArrayList<Integer> primeNumberList = new ArrayList<>();
public static void main(String[] args) throws NumberFormatException, IOException
{
BufferedReade... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
using namespace std;
bool p[1000000];
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
for (int i = 2; i < 1000000; i++){
p[i] = true;
}
for (int i = 2; i * i < 1000000; i++){
if (p[i]){
for (int j = 2 * i; j < 1000000; j += i){
p[j] = false;
}
}
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Arrays;
import java.util.Scanner;
public class Main
{
public static void main(String arg[])
{
Scanner sc = new Scanner(System.in);
while(sc.hasNext())
{
int n = sc.nextInt();
if(n==0)
return;
int ans =0;
boolean a[] = new boolean[1000001];
Arrays.fill(a, true);
a[0]=a[1... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | p = [True] * (105000)
p[0] = p[1] = False
for i in range(2, len(p)):
if p[i]:
for j in range(i * 2, len(p), i):
p[j] = False
p = [i for i in range(len(p)) if p[i]]
while True:
n = int(raw_input())
if n == 0:
break
print sum(p[:n]) | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
bool sosu[200000] = {false};
long long int ans[200000] = {0};
int main() {
vector<long long int> num;
for ( long long int i = 2; i < 200000; i++ ) {
ans[i] = ans[i-1];
if ( sosu[i] == false ) {
ans[i] += i;
num.push_back(i);
fo... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <cmath>
using namespace std;
bool flag[10000001];
int main(){
int n=0,ans=0,co=0;
for(int i=0;i <= 1000000;i++){
if(i<=1)flag[i]=false;
else flag[i]=true;
}
for(int i=2;i <= 1001 ;i++){
if(flag[i]==true){
for(int j=i*2 ; j<=1000000 ; j=j+i){
flag[j] = false;
}
}
}... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <stdio.h>
int main(){
int n,now=1,a[10000];
bool prime[1000000];
a[0]=2;
for(int i=3;i<1000000;i+=2){
prime[i]=true;
}
for(int i=3;now<10000;i+=2){
if(prime[i]){
a[now]=i+a[now-1];
now++;
if((long long)i*i<1000000){
for(int... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#include<vector>
using namespace std;
int main(){
vector<bool> prime(1000000,true);
vector<int> p;
for(int i = 2; i < 1000000; i++)if(prime[i]){
p.push_back(i);
for(int j = 2; i*j < 1000000; j++)prime[i*j] = false;
}
int n;
while(cin >> n,n){
int ans = ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys
def prime(m):
N=range(1,m+2,2)
r=int(m**.5)
h=len(N)
N[0]=0
for i in range(h):
x=N[i]
if x>r: break
if x and i+x<h: N[i+x:h:x]=[0] * ((h-1-i-x)/x+1)
N[0]=2
return filter(None,N)
PRIMES=prime(105000)
while 1:
n=input()
if n==0: break
print sum(PR... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <cstdio>
#define MAX 1000000
bool prime[MAX] = {true};
int p[10000] = {0};
void eratos(void);
int main(void)
{
eratos();
int n, sum;
while (scanf("%d", &n)){
if (n == 0) break;
sum = 0;
for (int i = 0; i < n; i++){
sum += p[i];
// printf("p[i] : %d\n", p[i]);
}
printf("%d\n", sum);... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys
n=15*10**4
p=[True]*n
p[0],p[1]=False,False
for i in xrange(2,int(n**0.5)+1):
if p[i]==True:
for j in xrange(i**2,n,i):
p[j]=False
s=[i for i in xrange(n) if p[i]]
while True:
n=input()
if n==0:break
print sum(s[0:n]) | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <algorithm>
#include <cassert>
#include <cctype>
#include <cstdio>
#include <math.h>
#include <map>
#include <queue>
#include <string>
using namespace std;
int prime[1000000],n;
bool isprime[1000001];
int sieve(int n){
int p=0;
for(int i=0;i<=n;i++)isprime[i]=true;
isprime[0]=isprime[1... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 |
import java.util.*;
import java.lang.*;
import java.math.*;
public class Main {
Scanner sc = new Scanner(System.in);
void run(){
boolean[] np = new boolean[500001];
np[0] = np[1] = true;
for(int i=0; i< 500001;i++){
if(np[i]) continue;
for(int j = i*2; j < 500001; j += i){
np[j] = true;... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Scanner;
class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static Scanner sc = new Scanner(new InputStreamReader(System.in));
public ... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys
n = 104729
a = [True] * n
a[0] = False
for i in range(2,n):
if(a[i-1]):
for j in range(i**2-1, n, i):
a[j] = False
a = [i[0]+1 for i in enumerate(a) if i[1]]
for s in sys.stdin:
n = int(s)
if n == 0:
exit()
print(sum(a[:n])) | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#include<vector>
#define MAX 999999
bool prime[MAX];
using namespace std;
int main(void){
int n ;
vector<int> dp(0,0);
//isPrime();
fill(prime,prime+MAX,true);
int cnt = 0;
prime[0]=prime[1]=false;
for(int i = 2 ; i < MAX ; i ++){
if(prime[i]){
cnt += i;
for(int j ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | //0053
#include<cstdio>
#include<algorithm>
int main(void)
{
const int N = 104729+1; //1000ÔÚÌfÍ104729
bool isPrime[N];
std::fill(isPrime, isPrime+N, true);
isPrime[0] = isPrime[1] = false;
int i, j;
for(i = 2; i*i <= N; i++)
{
if(isPrime[i])
{
for(j = i*2; j < N; j+=i)
{
isPrime[j] = fals... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(true){
int n=sc.nextInt();
if(n==0){
break;
}
if(n==1){
System.out.println(2);
break;
}
int count=1;
int sum=2;
for(int i=3;count<n;i+=2){
boolean f... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
int p[1000000];
int main()
{
int i,j,k=1;
vector<int> v(10000);
v[0] = 2;
for(i=3;i<1000000;i+=2){
if(p[i] == 0){
v[k] = i + v[k-1];
k++;
if(k >= 10000) break;
for(j=i*2;j<1000000;j+=i){
p[j] = 1;
}
}
}
while(c... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.*;
public class Main {
static boolean prime[] = new boolean[200000];
static int[] result = new int[10002];
static void Do(int k){
int sum=0;
for(int i=1;i<=k;i++){
sum+=P(i);
}
}
static int P(int n){
int cnt=0;
for(int i=0;i<prime.length;i++){
if(prime[i]){
cnt++;
if(cnt==n... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.math.BigInteger;
import java.util.*;
import static java.util.Arrays.*;
import static java.lang.Math.*;
// AOJ 0053
public class Main {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
sieve(1000000);
while (true) {
int n = sc.nextInt();
if (n == 0) break;
... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys,time
n = 104730
a = [True] * n
a[0] = a[1] = False
for i in range(2,n):
if(a[i]):
for j in range(i+i, n, i):
a[j] = False
a = [i[0] for i in enumerate(a) if i[1]]
for s in sys.stdin:
n = int(s)
if n == 0:
exit()
print(sum(a[:n])) | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<cstdio>
#include<cmath>
using namespace std;
int dp[10010];
int prime(int n) {
int i, j,d;
if (n == 1) { return dp[1] = 2; };
d = dp[n - 1];
for (i = d+1;; i++) {
for (j = 2;; j++) {
if (i%j == 0)break;
if (j == (int)sqrt(i)+1) { return dp[n] = i; }
}
}
}
int main() {
int n, i;
long long sum... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
int prime[10000];
int ptr = 0;
bool isPrime(int p)
{
if (p <= 1)return false;
for (int i = 2; i * i <= p; i++)
{
if (p % i == 0)return false;
}
return true;
}
void setPrime()
{
for (int i = 2;;i++)
{
if (ptr >= 10000)break;
if (isPrime(i))prime[ptr++] = i;
}
}
int ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <math.h>
#define N 1000000
using namespace std;
int main(){
int n;
int arr[N];
for(int i=0;i<N;i++){
arr[i]=1;
}
for(int i=2;i<sqrt(N);i++){
if(arr[i]){
for(int j=0;i*(j+2)<N;j++){
arr[i*(j+2)]=0;
}
}
}
while(true){
cin >> n;
if(n==0){
break;
}
int cnt=0... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 |
import java.util.*;
public class Main {
//f©Ç¤©ð»Ê·éÖ
private boolean isPrime(int n){
for(int i=2; i * i <= n; i++){
if(n % i == 0){
return false;
}
}
return true;
}
public void doIt(){
Scanner sc = new Scanner(System.in);
//GgXelXÌÓé¢
final int MAX = 999999;
b... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int[] v = new int[649900];
for(int i = 0;i < 649900;i++){
v[i] = 1;
}
int p = 3;
while(true){
if(p*p > 1299800){
break;
}else{
if(v[(p-1)/2] == 0){
p += 2;
... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
int prime[120001],num,cou=0,ans=0;
for(int i=2;i<=120000;i++)prime[i]=0;
for(int i=2;i<=(int)sqrt(120000);i++){
for(int j=2;j*i<=120000;j++)prime[i*j]=1;
}
while(cin>>num&&num!=0){
for(int i=2;cou<num;i++){
if(prim... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
int p[1<<17],n,s,i;
int main(){p[1]=1;for(i=2;i*i<1<<17;i++)if(!p[i])for(s=2;i*s<1<<17;s++)p[i*s]++;for(;std::cin>>n,n;){for(i=2,s=0;n;i++)if(!p[i])s+=i,n--;std::cout<<s<<"\n";}} | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#include<cstring>
int sumPrime[10000];
void makePrime(){
const int max=1000000;
bool sieve[max];
int k=0;
memset(sieve,true,sizeof(sieve));
for(int i=0;i<max&&k<10000;++i){
if(sieve[i]){
sumPrime[k]=i+2+(k>0?sumPrime[k-1]:0);
++k;
for(int j=i*2+2;j<max;j+=i+2){
sieve[j]=false;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | a=[True]*104743
p=[]
for i in range(2,int(104743**0.5)+1):
if a[i]:
for j in range(i*i,104743,i):a[j]=False
for i in range(2,104730):
if a[i]:p.append(i)
while 1:
n=int(input())
if n==0:break
print(sum(p[:n])) | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Calendar;
import java.util.Scanner;
public class Main {
Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
new Main().AOJ0053();
}
void AOJ0053(){
ArrayList<Integer> primes = primeTable(1000000);
while(true... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.Scanner;
class Main {
public static void main(String[] args) {
int[] prime = new int[10000000];
for(int i=0;i<10000000;i++) {
prime[i]=0;
}
for(int i=2; i<10000000; i++) {
if(prime[i]==1) {
continue;
}
prime[i]=2;
int j=i+i;
while(j<1000000... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
int p[104730]={0},n,s,i,j;
int main(){
p[1]=1;
for(i=2;i*i<104730;i++)
if(!p[i])
for(j=2;i*j<104730;j++)
p[i*j]=1;
while(cin>>n,n) {
for(i=2,s=0;n;i++)
if(!p[i])
s+=i,n--;
cout<<s<<endl;
}
} | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
using namespace std;
int isprime(int n){
if(n<2)return 0;
if(n==2)return 1;
if(n%2==0)return 0;
for(int i=3;i*i<=n;i+=2){
if(n%i==0)return 0;
}
return 1;
}
int main(){
int a,n[10001]={0},k;
k=0;
for(int i=1;k<=10000;i++){
if(isprime(i)){
k++;
n[k]=n[k-1]+i;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | def sieve(n):
primes = [True] * n
primes[0] = primes[1] = False
for i in xrange(2, int(n ** 0.5) + 1):
if primes[i]:
for j in xrange(i * i, n, i):
primes[j] = False
return [i for i in xrange(n) if primes[i]]
prime_list = sieve(200000)
while 1:
a = input()
if ... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#define x 200000
using namespace std;
int main(){
int p[x+1];
bool q[x+1];
for(int i=2;i<=x;i++) q[i] = true;
for(int i=2;i<=x;i++){
if(not q[i]) continue;
for(int j=2;j<=x/i;j++) q[i*j] = false;
}
int c = 0;
for(int i=2;i<=x;i++) if(q[i]) p[c++] = i;
int n;
while(1){
cin >> n;
if(... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#define N 104730
using namespace std;
int main(){
int n;
int m=0;
bool ip[N];
int p[10000];
for(int i=0;i<N;i++) ip[i]=true;
ip[0]=ip[1]=false;
for(int i=2;i<N;i++){
if(ip[i]){
p[m++]=i;
for(int j=2*i;j<N;j+=i) ip[j]=false;
}
}
while(1){
int a=0;
cin>>n... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
bool p[200000];
void erato(){
fill(p, p+200000, true);
p[0] = p[1] = false;
for(int i=0; i*i < 200000; i++){
if(!p[i]) continue;
for(int j= i+i; j < 200000; j += i){
p[j] = false;
}
}
}
int main(){
erato();
int ans, num, input;
while(cin >> input, input){
a... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | from itertools import *
n=range(104730);a=list(n)
for i in range(2,320):a[i*2::i]=[0]*len(a[i*2::i])
p=list(compress(n,a))
for e in iter(input,'0'):print(sum(p[:int(e)+1])-1)
| PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
#include<string>
#include<vector>
#include<cstdio>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<map>
#include<functional>
using namespace std;
int stoi(string x){stringstream ss;ss<<x;int tmp;ss>>tmp;return tmp;}
string itos(int x){stringstream ss;ss<<x;return ss.str();}
bool is(int... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | from itertools import accumulate, chain
ub = 105000
primes = {2, 3} | {m for n in (5, 7) for m in range(n, ub, 6)}
du = primes.difference_update
for n in chain(range(5, ub, 6), range(7, ub, 6)):
if n in primes:
du(range(n*3, ub, n*2))
cumsum = tuple(accumulate(primes))
while True:
n = int(input())
... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
int main() {
bool sosu[110000] = { false };
for (int i = 2; i <= sizeof(sosu) / sizeof(bool); ++i) {
if (sosu[i] == false) {
for (int k = i * 2; k <= sizeof(sosu) / sizeof(bool); k+=i) {
sosu[k] = true;
}
}
}
int answer;
int time;
while (1) {
cin >> time;
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
#define MAX_LIST 105000
int main(){
bool prime[MAX_LIST];
for(int i=0;i<MAX_LIST;i++){
prime[i]=1;
}
prime[0]=prime[1]=0;
for(int i=0;i*i<MAX_LIST;i++){
if(prime[i]==1){
for(int j=i*2;j<MAX_LIST;j=j+i){
prime[j]=0;
}
}
}
int primelist[10000];
int j=0;
for(int i=0;i<MAX_LIST;i++){... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <cstdio>
#include <bitset>
using namespace std;
#define MAX 1000001
bitset<MAX> bs;
int n;
long int s[10001] = { 0 };
int main(){
int it = 1;
for (int i = 2; i < MAX; i++) {
if (!bs[i]) {
for (int j = i + i; j < MAX; j += i) {
bs.set(j);
}
s[it] = s[it - 1] + i;
it++;
if (it > 10000) ... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <stdio.h>
#include <cmath>
#include <algorithm>
#define NUM 120000
int main(){
int table[NUM],limit;
for(int i=0; i < NUM;i++)table[i] = 1;
table[0] = 0;
table[1] = 0;
limit = sqrt(NUM);
for(int i=2;i<=limit;i++){
if(table[i] == 1){
for(int k=2*i;k < NUM; k += i){
table[k] = 0;
}
}
}
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
int main(void){
bool prime[1000000];
for(int i=0;i<=1000000;i++)
prime[i]=true;
prime[0]=prime[1]=false;
for(int i=2;i<=1000000;i++)
if(prime[i])
for(int j=2*i;j<=1000000;j+=i)
prime[j]=false;
int sum[10001]={0};
int index=1;
for(int i=0;i<=1000000;i++){
if(... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
int num[500000]={0};
int main(){
for(int i=2;i<sqrt(500000);i++){
for(int j=i;j<500000;j+=i){
if(num[j]==0 && j!=i){
num[j]=1;
}
}
}
while(1){
int n;
cin>>n;
if(n==0) break;
int cnt=0;
long long sum=0;
for(int i=2;i<500000;i... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <stdio.h>
#include <math.h>
int prime[200000];
void Eratos(void);
int main(void){
int n;
int p[10000];
int num = 0;
int ans;
Eratos();
for(int i=0; num < 10000; i++){
if(prime[i] == 1){
p[num] = i;
num++;
}
}
while(1){
ans = 0;
scanf("%d", &n);
if(n == 0) break;
for(int i=0; i<... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<iostream>
using namespace std;
int main(){
int p[10000];
int j=3;
int a;
bool c;
p[0]=2;
for(int i=1;i<10000;i++){
for(;;j++){
c=true;
for(int k=2;k*k<=j;k++){
a=j;
a%=k;
if(a==0){c=false; break;}}
if(c){
p[i]=p[i-1]+j;
j++;
break;}
}}
int n;
while(cin>>n){
if(n==0)break;
cout<<p[n-1]<<endl;}} | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import sys
def prime(m):
N=range(1,m+2,2)
r=int(m**.5)
h=len(N)
N[0]=0
for i in range(h):
x=N[i]
if x>r: break
if x and i+x<h: N[i+x:h:x]=[0] * ((h-1-i-x)/x+1)
N[0]=2
return filter(None,N)
PRIMES=prime(105000)
while 1:
n = input()
if n==0: break
s = sum(PR... | PYTHON |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<stdio.h>
int main(void){
int i,j,k;
int prime[10000],sum[10000];
int n;
i = 1;
k = 3;
prime[0] = sum[0] = 2;
while(i<10000){
n = 1;
for(j=0;prime[j]*prime[j]<=k;j++){
if(!(k%prime[j])){
n = 0;
break;
}
}
if(n){
prime[i] = k;
sum[i] = sum[i-1]+k;
i... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
ArrayList<Integer> primes = prime(104740);
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int n = sc.nextInt();
if (n == 0) {
break;
}
long sum = 0;
... | JAVA |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
long long int prime[120001],num,cou=0,ans=0;
for(int i=2;i<=120000;i++)prime[i]=0;
for(int i=2;i<=(int)sqrt(120000);i++){
for(int j=2;j*i<=120000;j++)prime[i*j]=1;
}
while(cin>>num&&num!=0){
for(int i=2;cou<num;i++){
... | CPP |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | import math
M = 104729
lst = list(range(M+1))
for i in range(0, M+1, 2):
lst[i] = 0
lst[1] = 0
lst[2] = 2
for i in range(3, math.ceil(math.sqrt(M+1)), 2):
if lst[i]:
for j in range(i**2, M+1, i):
lst[j] = 0
anslst = [i for i in lst if i != 0]
while True:
n = int(input())
if n ==... | PYTHON3 |
p00053 Sum of Prime Numbers | Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 ... | 7 | 0 | #include <iostream>
#include <vector>
using namespace std;
#define MAX 1000000
vector<int> prime;
bool is_prime[MAX+1];
void Sieve() {
for(int i=0; i<=MAX; i++) {
is_prime[i] = true;
}
is_prime[0] = is_prime[1] = false;
for(int i=2; i<=MAX; i++) {
if(is_prime[i]) {
prime.push_back( i )... | CPP |
p00183 Black-and-White | Consider a tic-tac-toe on a 3x3 board. Tic-tac-toe is a two-player battle game. Decide the first attack and the second attack, one hits Kuroishi and one hits Shiraishi. The winner is the person who puts stones one by one on the board alternately and arranges three of his own stones first in either the vertical, horizon... | 7 | 0 | #include <iostream>
#include <string>
using namespace std;
int main(){
char d[9];
string m[8]={"012","345","678","036","147","258","048","246"};
int f;
while(cin>>d[0],d[0]!='0'){
for(int i=1;i<=8;i++)cin>>d[i];
f=1;
for(int i=0;i<8;i++){
if(d[m[i][0]-'0']==d[m[i][1]-'0']&&d[m[i][1]-'0']==d[m[i][2]-'0']&&... | CPP |
p00183 Black-and-White | Consider a tic-tac-toe on a 3x3 board. Tic-tac-toe is a two-player battle game. Decide the first attack and the second attack, one hits Kuroishi and one hits Shiraishi. The winner is the person who puts stones one by one on the board alternately and arranges three of his own stones first in either the vertical, horizon... | 7 | 0 | #include<iostream>
#include<string>
using namespace std;
int main(){
while(true){
string brd[3];
bool bWin=false;
bool wWin=false;
for(int i = 0; i < 3; ++i){
cin >> brd[i];
if( brd[i] == "0") return 0;
}
// row
for(int i = 0; i < 3; ++i){
int b=0,w=0;
for(int... | CPP |
p00183 Black-and-White | Consider a tic-tac-toe on a 3x3 board. Tic-tac-toe is a two-player battle game. Decide the first attack and the second attack, one hits Kuroishi and one hits Shiraishi. The winner is the person who puts stones one by one on the board alternately and arranges three of his own stones first in either the vertical, horizon... | 7 | 0 | #include <iostream>
#include <cstdio>
using namespace std;
string f[3];
int main() {
while (cin>>f[0]) {
if (f[0]=="0") break;
for (int i=1; i<3; i++) cin>>f[i];
char res='+';
for (int i=0; i<3; i++) {
if (f[0][i]==f[1][i]&&f[1][i]==f[2][i]) {
if (f[0][i]!... | CPP |
p00183 Black-and-White | Consider a tic-tac-toe on a 3x3 board. Tic-tac-toe is a two-player battle game. Decide the first attack and the second attack, one hits Kuroishi and one hits Shiraishi. The winner is the person who puts stones one by one on the board alternately and arranges three of his own stones first in either the vertical, horizon... | 7 | 0 | #include<iostream>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>
using namespace std;
typedef long long ll;
typedef unsigned long ... | CPP |
p00183 Black-and-White | Consider a tic-tac-toe on a 3x3 board. Tic-tac-toe is a two-player battle game. Decide the first attack and the second attack, one hits Kuroishi and one hits Shiraishi. The winner is the person who puts stones one by one on the board alternately and arranges three of his own stones first in either the vertical, horizon... | 7 | 0 | chk = [
(1,2,3),
(4,5,6),
(7,8,9),
(1,4,7),
(2,5,8),
(3,6,9),
(1,5,9),
(3,5,7)
]
while True:
s = "0"
s += raw_input()
if s[1]=='0':
break
s += raw_input()+raw_input()
judge = "NA"
for ch in chk:
if s[ch[0]]==s[ch[1]]==s[ch[2]]:
if s... | PYTHON |
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