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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 | //Sum of Prime Numbers
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> p;
int main(){
p.push_back(2);
for(int i=3; p.size()!=10000; i+=2){
bool flag=true;
for(int j=0; j<p.size(); j++){
if(i%p[j]==0){
flag=false;
break;
}
}
if(flag)p.push_b... | 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 <cstring>
#define N 110000
using namespace std;
int main(void){
int p[N] = {0};
int sum[10001]={0}, cnt=0;
for (int i=2; i<N; i++){
if (p[i]==-1) continue;
p[i]=0;
cnt++;
sum[cnt] = sum[cnt-1]+i;
for (int j=i; j<N; j+=i){
p[j]=-1;
}
}
while (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 | #include<iostream>
#include<vector>
using namespace std;
inline void prime(int *p, int n){
fill(p, p+n, 1);
p[0] = 0;
p[1] = 0;
for(int i=2; i<=n; i++){
for(int j=2; i*j<=n; j++){
p[i*j] = 0;
}
}
}
int main(){
int cnt, p[105001], sum, n;
vector<int> vc;
prime(p, 105000);
cnt = 0;
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>
#include <vector>
#include <cmath>
using namespace std;
int main() {
int n;
vector<int> pri(1000000, 1);
vector<int> p_sum;
pri[0] = 0;
pri[1] = 0;
for(int i = 2; i < sqrt(1000000); i++) {
for(int j = i*i; j < 1000000; j+=i) {
pri[j] = 0;
}
}
int cnt = 0;
p_sum.pus... | 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 <cmath>
#include <iostream>
#include <vector>
typedef std::vector <int> vi;
vi PRIMES;
vi prime(int n){
int i=0,j=1,h=(n+1)/2,x;
vi s(h),p;
while(j<=n){
s[i++]=j;
j+=2;
}
for (i=1;i<=(int)pow(n,0.5);i++)
if (x=s[i]) for (j=i+x;j<h;j+=x) s[j]=0;
s[0]=2;
for (i=0;i<h;i++) if(s[i]!=0) p.push_back(s[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 | Max=1000000
Sqrt=1000
def prime_number():
count=[]
prime=[True for i in range(Max)]
for i in range(4,Max,2):
prime[i]=False
for i in range(3,Sqrt,2):
if prime[i]==True:
for j in range(i*2,Max,i):
prime[j]=False
for k in range(2,Max):
if prim... | 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(){
const int NUM = 1e6;
bool prime[NUM];
for(int i=2;i<NUM;i++)
prime[i] = true;
for(int i=2;i*i<NUM;i++)
for(int k=2;i*k<NUM;k++)
prime[i*k]=false;
int n,c;
while(cin>>n){
if(!n)break;c=0;
for(int i=2,k=0;k<n;i++)if(prime[i])c+=i,k++;... | 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;
bool isprime(int n) {
for (int i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return n != 1;
}
int main() {
vector<long long> v;
int n = 0;
for (int i = 2; n <= 10000; i++) {
if (isprime(i)) {
v.push_back(i);
n++;
}
}
while (cin ... | 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 | primes = [0, 0] + [1] * 150000
for i in range(2, 388):
if primes[i]:
for j in range(i*i, 150001, i):
primes[j] = 0
values = [i for i, v in enumerate(primes) if v]
while True:
n = int(input())
if n == 0:
break
print(sum(values[: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>
#define N 300000
using namespace std;
bool p[N];
int num[N];
int main(){
for(int i=0;i<N;i++)p[i] = true;
p[0] = p[1] = false;
for(int i=2;i*i<N;i++){
if(p[i]){
for(int j=2*i;j<N;j+=i)p[j] = false;
}
}
vector<int> v;
int sum = 0,pos = 0;
while(v.size... | 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 ans[10001],size=2;
bool isP(int x){
if(x<=1)return false;
for(int i=2;i*i<=x;i++){
if(x%i==0)return false;
}
return true;
}
int main(){
ans[1]=2;
for(int i=3;size<10001;i++){
if(isP(i)){
ans[size]=ans[size-1]+i;
size++;
}
}
int n;
whi... | 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 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
ArrayList<Integer> index = new ArrayList<Integer>();
ArrayList<Integer> search = new ArrayList<Integer>();
for(int i=2; i<=110000; i++) search.add(i);
index.add(2);
while(index.get(index.s... | 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
PRIMES=[]
SIEVES=[]
def sieve(max_number):
numbers = range(1, max_number + 2, 2)
mroot = int(max_number ** 0.5)+1
half = len(numbers)
numbers[0]=0
for i in range(0, half):
x = numbers[i]
if x > mroot: break
if x and i + x < half:
numbers[i + x: half: x... | 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>
using namespace std;
int main(){
bool num[1000001] = {false};
for(int i=2; i<1000; i++) if(!num[i]) for(int j=i*i; j<1000001; j+=i) num[j] = true;
int n;
while(cin >>n,n){
int ans = 0;
for(int i=2; n>0; i++) if(!num[i]){ans+=i;n--;}
cout <<ans<<endl;
}
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 | #include <stdio.h>
#define NN 999999
int main(){
int i,j,n,sum;
bool prime[NN];
for(i=0;i<NN;i++) prime[i]=true;
while(1){
sum=0;n=0;
scanf("%d",&n);
if(!n) break;
for(i=2;i<NN;i++){
for(j=i*2;j<NN;j=j+i) prime[j]=false;
}
for(i=2,j=0;j<n;i++){
if(prime[i]==true){
sum+=i;
j++;
}
}
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 | def isPrime(x):
if x == 2:
return True
if x < 2 or x % 2 == 0:
return False
i, root_x = 3, int(pow(x, 0.5))
while i <= root_x:
if x % i == 0:
return False
i += 2
return True
primes = [2]
for i in range(3, 104730):
if isPrime(i):
primes.append(... | 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 is_prime[1000000];
is_prime[0]=is_prime[1]=false;
for(int i=2;i<1000000;i++){
is_prime[i]=true;
}
for(int i=0;i*i<1000000;i++){
if(is_prime[i]){
for(int j=i*i;j<1000000;j+=i){
is_prime[j]=false;
}
}
}
while(1){
int a=0;
cin>>a;
int cu... | 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<string.h>
#include<math.h>
const int max=104729;
char so[104730];
int main(){
int i,j,n,s;
memset(so,0,sizeof(so));
so[0]=so[1]=1;
for(i=2;i*i<=max;i++){
for(j=2*i;j<=max;j+=i){
so[j]=1;
}
}
while(0<=scanf("%d",&n)){
if(n==0)break;
s=0;
for(i=1;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>
#define MAX 1000000
using namespace std;
int main(){
bool p[MAX];
p[0] = p[1] = false;
for(int i=2;i<MAX;i++)p[i] = true;
for(int i=2;i<MAX;i++){
if(p[i]){
for(int j=i+i;j<MAX;j=j+i){
p[j] = false;
}
}
}
int n;
while (cin >> n,n){
int sum = 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 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] pri = new int[260000];
for(int i=2;i<=250000;i++){
pri[i] = i;
}
for(int i=2;i<=300000;i++){
for(int j=i+i;j<=250000;j+=i){
pri[j]=0;
}
}
while(true){
int 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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class Main {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while(true){
int numOfPrimeNumbers = Integer.parseInt(br.readLine()... | 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;
using db = double;
using ll = long long;
using vi = vector <int>;
#define op operator
#define pb push_back
const int N = 2e5L + 11;
int p[N];
int main() {
iota(p, p + N, 0);
for(int i = 2; i < N; i ++) if(p[i] == i)
for(int j = i; j < N; j += i) p[j] = i;
ios :: sy... | 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(110000, false);
prime[2] = true;
for (int i = 3; i < 110000; i += 2) {
prime[i] = true;
}
for (int i = 3; i < 110000; i += 2) {
for (int j = i + i; j <= 110000; j += i) {
prime[j] = false;
}
}
int n;
while (cin... | 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 <cstring>
using namespace std;
const int PMAX = 120000;
int main() {
bool isnp[PMAX+1];
int p[10000];
int x;
memset(isnp, 0, sizeof(bool)*(PMAX+1));
for (int i=2; i*i<=PMAX; i++) {
if (!isnp[i]) {
for (int j=2*i; j<=PMAX; j+=i) {
isnp[j] = true;
}
}
}... | 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>
#include<cmath>
using namespace std;
int main(){
vector<int>pri;
pri.push_back(2);
for(int i=3;;i+=2){
int flag=1;
for(int j=2;j<=sqrt(i);j++){
if(i%j==0){
flag=0;
break;
}
}
if(flag)
pri.push_back(i);
if(pri.size()==10000)
break;
... | 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
prime = [0 for i in range(11000)]
k = 0
r = 104900
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 | def sieve(n):
a = range(n)
a[:2] = None, None
for i in range(2, n):
if i ** 2 >= n: break
if not a[i]: continue
for i in range(i ** 2, n, i):
a[i] = None
return [v for v in a if v]
def partialsum(a):
b = [0] + a
for i in range(len(a)):
b[i + 1] += b[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 <stdio.h>
using namespace std;
int f[110000],pr[11000];
int main() {
int n=0,i,j,s=0;
for (i=2;i<110000;i++) {
if (f[i]==0) { s+=i; pr[n++]=s; for (j=i;j<110000;j+=i) f[j]=1;}
}
while(cin >> n) {
if (n==0) break;
cout << pr[n-1] << endl;
}
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 | def prime_table(n):
list = [True for _ in xrange(n + 1)]
i = 2
while i * i <= n:
if list[i]:
j = i + i
while j <= n:
list[j] = False
j += i
i += 1
table = [i for i in xrange(n + 1) if list[i] and i >= 2]
return table
table = prime_table(105000)
while True:
n = int(raw... | 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 <vector>
using namespace std;
int main()
{
vector<int> prime;
int n;
prime.push_back(2);
for(int i = 3; prime.size() < 10000; i += 2){
for (int j = 0; j < prime.size() ;j++){
if (i%prime[j] == 0){
break;
}
if (j == prime.size()-1){
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 | s = [0]
M = 104850
p = [1]*M
S = 0
for i in range(2,M):
if p[i]:
for j in range(i*i,M,i):
p[j] = 0
S += i; s.append(S)
while 1:
n = input()
if n==0: break
print s[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 | import sys
PRIMES=[]
SIEVES=[]
def sieve(m):
N=range(1,m+2,2)
r=int(m**.5)+1
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)
return N
def prime(n):
global SIEVES
if SIEVES==[]: SIEVES=sieve(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<bits/stdc++.h>
using namespace std;
int P[10010];
int sum[10010]={0};
void prime() {
int i=3,j,k,cnt=1;
P[0] = 2;
while(cnt < 10005){
k=0;
for(j=3;j<=sqrt(i);j+=2){
if(i%j==0){
k=1;
break;
}
}
if(k==0) {
P[cnt] = i;
cnt++;
}
i+=2;
}
sum[0] = P[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>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int j;
int prime(){
int k,fg=0;
while(1){
if(j%2==0){
fg=1;
j++;
}
else{
k=3;
while(k<=sqrt(j)){
if(j%k==0){
fg=1;
j++;
break;
}
k+=2;
}
}
if(fg==0){
j++;
return j-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.io.*;
import java.awt.geom.*;
import java.math.*;
public class Main {
static final Scanner in = new Scanner(System.in);
static final PrintWriter out = new PrintWriter(System.out,false);
static ArrayList<Integer> ps;
static boolean solve() {
int n = in.nextInt();
if (n == 0) r... | 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 main(){
int dt[150000];
dt[0]=0;
dt[1]=0;
for(int i=2;i<150000;i++){
dt[i]=1;
}
for(int i=2;i<750;i++){
for(int j=2*i;j<150000;j+=i){
dt[j]=0;
}
}
vector<int> prime;
for(int i=0;prime.size()<10001;i++){
if(dt[i]==1){
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 | #include<iostream>
#include<cmath>
using namespace std;
int x[1000000], y[100000], n, cnt;
int main() {
for (int i = 2; i < 200000; i++) {
for (int j = 2; j <= sqrt(i); j++) {
if (i%j == 0) { goto E; }
}
x[i] = 1;
E:;
}
for (int i = 0; i < 200000; i++) {
if (x[i] == 1) { cnt++; y[cnt] = y[cnt - 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 | #include <stdio.h>
int a[2000000];
int main(){
int n;
for(int i = 0; i < 2000000; i++){
a[i] = 1;
}
for(int i = 2; i < 2000000; i++){
if(a[i]){
for(int j = i * 2;j < 2000000; j = j + i){
a[j] = 0;
}
}
}
scanf("%d",&n);
while(n != 0){
long int ans = 0;
for(int i = 2; i < 2000000; i++){
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>
#include<cmath>
using namespace std;
bool memo[104730]={false};
void prime(void){
bool flag;
memo[2]=true;
for(int i=3;i<104730;i=i+2){
flag=true;
for(int j=3;j<=sqrt((double)i);j=j+2){
if(i%j==0){
flag=false;
break;
}
}
if(flag==true){
memo[i]=true;
}
}
return;
... | 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 <cstdlib>
#include <numeric>
#include <vector>
using namespace std;
vector<int> sieve(int n) {
vector<int> res(1, 2);
vector<bool> prime(n, true);
for(int i = 3; i < n; i += 2) {
if(prime[i]) {
res.push_back(i);
for(int j = i * 3; j < n; j += i * 2)
prime[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<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
ll prime[1000001];
ll length = 0;
void insert(ll p) { prime[length] = p; length++; return; }
int main() {
insert(2);
for (ll i = 3; i < 1000000; i += 2) {
for (ll j = 0; j < length; j++)
if (i%prime[j] == 0)goto cont;
else if (prime... | 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.*;
import java.util.*;
class Main{
public static void main(String[] args){
BufferedReader sc=new BufferedReader(new InputStreamReader(System.in));
LinkedList<Integer> list = new LinkedList<Integer>();
boolean[] prime = new boolean[110000];
int[] s = new int[10000];
prime[0] = true; prime[1] = t... | 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 <cmath>
#include <iostream>
#include <vector>
typedef std::vector <int> vi;
vi PRIMES;
vi prime(int n){
vi s,p;
int i,j,x,h;
for (i=1;i<=n;i+=2) s.push_back(i);
h=s.size();
s[0]=0;
for (i=1;i<=(int)pow(n,0.5);i++)
if (x=s[i]) for (j=i+x;j<h;j+=x) s[j]=0;
s[0]=2;
for (i=0;i<h;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.Scanner;
public class Main {
public static void main(String[] args){
Scanner scan = null;
try{
int num;
int[] prime = new int[10000];
prime[0] = 2;
for(int i = 3 ,j = 1;j <prime.length;i += 2){
boolean check = false;
for(int k = 3; k < i /2; k += 2){
if(i % k ==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 | //Volume0-0053
import java.util.Scanner;
public class Main {
//declare
private static final int PRM_MAX = 104730,
PRM_NUM = 10001;
private static boolean [] prm_table = new boolean[PRM_MAX];
private static int [] prm_sum = new int [PRM_NUM];
... | 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 = 104730
a = [True] * n
a[0] = a[1] = False
for i in xrange(2,n):
if(a[i]):
for j in xrange(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 <cmath>
#include <iostream>
#include <vector>
typedef std::vector <int> vi;
vi P;
vi prime(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... | 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 | public class Main{
static int N = 1000001;
public void run(java.io.InputStream in, java.io.PrintStream out){
java.util.Scanner sc = new java.util.Scanner(in);
int[] p;
int pn, i, n, sum;
p = new int[N / 6];
pn = getprime(p);
for(;;){
n = sc.nextInt();
if(n == 0)break;
sum... | 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<stdio.h>
int table[500000];
int prime[100000];
int main(){
table[0]=-1;
table[1]=-1;
for(int i=2;i<500000;i++){
if(table[i]!=-1){
table[i]=1;
for(int j=i*2;j<500000;j+=i){
table[j]=-1;
}
}
}
int now=0;
for(int i=0;i<500... | 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
f = sys.stdin
def create_prime(n):
prime = [1] * (n + 1)
prime[:2] = [0, 0]
for i in range(len(prime)):
if prime[i]:
for j in range(2 * i, len(prime), i):
prime[j] = 0
return prime
prime = create_prime(200000)
while True:
n = int(f.readline())
i... | 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.*;
public class Main{
static Scanner kbd = new Scanner(System.in);
public static void main(String[] args){
int[] p = new int[1000000];
setP(p);
while(kbd.hasNext()){
int n = kbd.nextInt();
if(n!=0) solve(p, n);
}
}
static void solve(int[] p, int n){
int sum=0, i;
for(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 java.util.ArrayList;
import java.util.Scanner;
public class Main {
static ArrayList<Integer> list=new ArrayList<>();
static boolean[] prime=new boolean[104730];
static void eratos() {
for(int i=2; i<104730; i++) {
prime[i]=true;
}
for(int i=2; i<104730; i++) {
if(prime[i]) {
list.add(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 java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int[] sum = new int[10001];
int number = 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 | #include <iostream>
using namespace std;
int main()
{
int num[104730] = {0}, Pnum[10000], n, ans;
int k = 0;
for (int i = 2; i < 104730; i++) {
if (!(num[i])) {
Pnum[k] = i;
k++;
for (int j = i + i; j < 104730; j+=i) {
num[j] = 1;
}
}
}
while (cin >> n) {
if (n == 0) break;
ans = 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 | 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))
print('\n'.join(str(sum(p[:int(e)]))for e in iter(input,'0')))
| 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 itertools
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(itertools.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 | #include <iostream>
using namespace std;
bool t[999999] = {false};
int main()
{
int n, c, i, j, d;
t[0] = true;
for (i = 2; i * i <= 999999; i++)
{
if (!t[i-1])
{
for (j = i+i; j <= 999999; j+=i)
t[j-1] = true;
}
}
while (cin >> n, n)
{
c = 0;
d = 1;
for (i = 1; d <= n; i++)
{
if (!t[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>
#include <vector>
using namespace std;
#define N 104730
bool table[N];
vector<int> prime;
int main() {
int size = 0;
for (int i = 2; i < N; i++) {
if (table[i] == false) {
for (int j = i; j < N; j += i) {
table[j] = true;
}
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>
#include<vector>
#define loop(i,a,b) for(long long i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
using namespace std;
bool prime[7000000];
int main(){
rep(i,7000000)prime[i]=true;
prime[0]=prime[1]=false;
loop(i,2,10000){
for(long long j=i*2;j<7000000;j+=i)prime[j]=false;
}
vector<long long> 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 | def mark(s, x):
for i in range(x + x, len(s), x):
s[i] = False
def sieve(n):
s = [True] * n
for x in range(2, int(n**0.5) + 1):
if s[x]: mark(s, x)
return [i for i in range(0,n) if s[i] and i > 1]
n = []
while True:
tmp = int(input())
if tmp == 0:
break
n.append(tmp... | 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 <cstdio>
using namespace std;
#define rep2(x,from,to) for(int x = (from); x < (to); ++(x))
#define rep(x,to) rep2(x,0,to)
#define MAX_PRIME 160000
int main() {
bool p[MAX_PRIME];
p[0] = p[1] = 0;
rep2(i,2,MAX_PRIME) {
p[i] = 1;
}
rep(i,400) {
if(p[i]) {
for(int j = 2; i * j < MA... | 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 <cmath>
#include <iostream>
using namespace std;
int a[20000];
int sum[20000];
int n,s;
int main() {
a[0] = 2;
sum[0] = 2;
cin >> n;
while(n!=0) {
for(int i=1 ; i<n; ++i) {
int k = a[i-1] + 1;
while(a[i]==0) {
while(k%a[s]!=0) {
if(a[s]*a[s] > k) {
a[i] = k;
//cout << i+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.Arrays;
import java.util.List;
import java.util.Scanner;
public class Main{
static int max = 2000000;
static boolean[] isPrime = new boolean[max];
public static void main(String[] args) {
aryPrime();
List<Integer> list = new ArrayList<Integer>();
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 java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws NumberFormatException,
IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(
System.in));
StringBuilder builder = new Str... | 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
def pri(n):
if n < 2:
return False
elif n == 2:
return True
elif n % 2 == 0:
return False
i = 3
while i <= math.sqrt(n):
if n % i == 0:
return False
i += 2
return True
l=[i for i in range(110000)if pri(i) ]
while True:
n = in... | 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;
int main() {
const long N = 1000000;
vector<bool> p(N, true); p[0] = p[1] = false;
int i, j, n, a, c;
for (i = 2; i * i < N; i++) {
if (!p[i]) continue;
for (j = i * i; j < N; j += i) p[j] = false;
}
while (cin >> n, n) {
a = 0, c = 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 {
public static void main(String[] args) throws java.io.IOException{
Scanner scan = new Scanner(System.in);
while(true){
int n=scan.nextInt();
if(n==0)break;
int count=0;
int sum=0;
for(in... | 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) {
Scanner sc=new Scanner(System.in);
int input=0;
int sosuko=0;
long wa=0l;
int genzai=0;
boolean ss=false;
while(sc.hasNext()) {
input=sc.nextInt();
sosuko=0;
wa=0l;
genzai=0;
if(input>=1) {
wa+=2;
genz... | 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>
#include<vector>
using namespace std;
int main(){
int n;
vector<bool> ar(1000*1000+1,true);
for(int i=2;i<=1000;i++){
if(ar[i])for(int j=i*i;j<ar.size();j+=i)ar[j]=false;
}
vector<long> ps;
for(int i=2;i<ar.size();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 <iostream>
using namespace std;
/** Problem0044 : Prime Number II **/
#define MAX 110000
int main()
{
bool prime[MAX];
int input;
unsigned sum, num;
for (int i=0; i<MAX; i++)
prime[i] = true;
prime[0] = false; prime[1] = false;
for (int i=2; i<MAX; i++) {
if (prime[i]) {
for (int j=2; 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<stdio.h>
#include<math.h>
int main(){
bool N[10002];
//for(int i=0;i<10001;i++)
//N[i]=false;
int R[10003]={};
int H=2;
for(int i=3;H<=10001;i+=2)
{bool G=true;
for(int j=2;j<=sqrt((double)i);j++)
{if(i%j==0){G=false;break;}}
if(G==true){R[H]=i;H++;}
}
//N[2]=true;N[3]=true;
R[1]=2;R[0]=0;
int n;
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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Main {
/**
* @param args
* @throws IOException
* @throws NumberFormatException
*/
public static void main(String[] args) throws NumberFormatException, IOException {
// TODO... | 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.InputStreamReader;
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = null;
// エラトステネスの篩
int MAX = 1000000;
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 java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
static final int MAX_NUM_OF_PRIME_NUMBERS = 10000;
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while(true){
int numOfPrimeNumbers = Int... | 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;
#define MAX 10005
#define MAXL 1000000
//const MAX = 10001;
int ans[MAX];
int main(){
int n;
static bool c[MAXL]={true, true};
for(int i = 2;i < MAXL;i++){
if(!c[i]){
for(int j = 2;i*j < MAXL;j++)c[i*j] = true;
}
}
int j = 1;
for(int i = 1;j < MAX;i++){
if(!c[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>
using namespace std;
bool prime[500000];
void check(){
for(int i = 3; i < 400000; i+=2){
if(prime[i]){
for(int j = i*2; j< 400000; j+=i){
prime[j] = false;
}
}
}
}
int main(void){
int n;
for(int i = 3; i < 400000; i+=2){
prime[i] = true;
}
prime[2] = true;
check();
whi... | 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.io.*;
public class Main{
public static void main(String[] args){
Scanner stdIn = new Scanner(System.in);
Map<Integer, Integer> prime = new LinkedHashMap<Integer, Integer>();
List<Integer> result = new ArrayList<Integer>();
prime.put(1,2); prime.put(2,5);
int[] p = new int[... | 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 sc = new Scanner(System.in);
boolean[] p = new boolean[105001];
for (int i = 0; i <= 105000; i++) {
p[i] = true;
}
for (int i = 2; i <= 325; i++) {
for (int j = i * 2; j <= 105000; j += i) {
p[j] = false... | 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 main() {
int data[150000] = {};
int size = sizeof(data) / sizeof(int);
for (int i = 2; i < size; ++i) {
for (int j = i * 2; j < size; j += i)
data[j] = 1;
}
while (true) {
int n;
cin >> n;
if (n == 0)
break;
int ans = 0;
for... | 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>
using namespace std;
int main() {
bool prime[500000] = { false };
for (int i = 2; i <= sqrt(500000); i++) {
if (!prime[i]) {
for (int j = 2; j*i <= 500000; j++) {
prime[j*i] = true;
}
}
}
int n; unsigned int sum;
while (cin >> n) {
if (n == 0)break;
sum = 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 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
class Main{
static boolean[] ints=new boolean[104729+1];
static int[] primes=new int[10000+1];
static void makePrimes(){//System.out.println("ee");
ints[0]=true;ints[1]=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 | #include<iostream>
using namespace std;
int main(){
bool num[1000000] = {0};
for(int i=2; i<1000; i++){
if(!num[i]){
for(int j=i*i; j<1000000; j+=i){num[j] = 1;}
}
}
for(;;){
int n,sum = 0;
cin >>n;
if(n == 0){break;}
for(int i=2; n>0; i++){
if(!num[i]){n--;sum+=i;}
}
cout <<sum<<endl;
}
ret... | 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 prime(int x){
for(int i=2;i<=sqrt((double)x);i++){
if(x%i==0)return false;
}
return true;
}
int main(){
int n;
while(cin>>n,n){
int sum=0;
for(int i=2;n>=1;i++){
if(prime(i)){sum+=i;n--;}
}
cout<<sum<<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 | //Volume0-0053
import java.util.Scanner;
public class Main {
//declare
private static final int PRM_MAX = 104730,
PRM_NUM = 10001;
private static boolean [] prm_table = new boolean[PRM_MAX];
private static int [] prm_sum = new int [PRM_NUM];
... | 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 get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
MAX = 105000
primes = list()
for i in range(MAX):
primes.append(True)
primes[0] = False
primes[1] = False
for i in range(2, MAX):
j = i + i
while j < MAX:
primes[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 | def prime(t):
i=2
while i**2<=t:
if t%i==0:
return 0
i+=1
return 1
l=[]
for i in range(2,104740):
if prime(i):
l.append(i)
while 1:
n=int(raw_input())
if n==0:
break
ans=0
while n>0:
n-=1
ans+=l[n]
print ans | 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>
using namespace std;
int main(){
bool num[1000000]={true,true};
for(int i=2; i<1000; i++)
if(!num[i])
for(int j=i*i; j<1000000; j+=i)
num[j] = true;
int n;
while(cin >>n,n){
int ans = 0;
for(int i=2; n>0; i++){
if(!num[i]){
ans+=i;
n--;
}
}
cout <<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 | #include<iostream>
#include<cmath>
#define NUM 1000000
using namespace std;
int main(){
int n, prime[NUM+1], p[NUM+1], count, sum;
prime[1]=0;
for(int i=2; i<NUM+1; i++){
prime[i] = 1;
p[i-2]=0;
}
for(int i=2; i<=NUM/2; i++){
for(int j=2; i*j <= NUM; j++){
prime[i*j]=0;
}
}
co... | 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>
bool prime[104730];
int main() {
for(int i=2;i*i<=104729;i++)
for(int j=i*2;j<=104729;j+=i)prime[j]=true;
int n;
while(scanf("%d",&n),n) {
int p=0,sum=0;
for(int i=2;i<=104729;i++) {
if(!prime[i])sum+=i,p++;
if(p==n)break;
}
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 | #include <iostream>
#include <vector>
using namespace std;
typedef unsigned long long int lli;
vector<lli> p;
int main(void){
bool ps[200000];
for(int i=0;i<200000;i++)ps[i]=true;
ps[0]=ps[1]=false;
p.push_back(0);
for(int i=2;i<200000;i++){
if(ps[i]){
for(int j=2*i;j<200000;j+=i) ps[j]=false;
p.push_back... | 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;
#define MAX 10000000
typedef long long ll;
bool prime[MAX];
void calcPrime(){
for( ll i = 2; i < MAX; i++ ) prime[i] = true;
for( ll i = 2; i < MAX; i++ ){
if( prime[i] ){
for( ll j = i+i; j < MAX; j += i ) prime[j] = false;
}
}
}
int main(){
calc... | 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
def pri(n):
l =[True] * n
l[0] = l[1] = False
for i in range(2,n):
if l[i]:
for j in range(i*2, n, i):
l[j] = False
return l
n=105000
s=pri(n)
l=[i for i in range(n) if s[i]]
while True:
n = int(input())
if n == 0: break
print(sum(l[: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<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 | #define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"
using namespace std;
int const N = 10010;
int ans[N];
bool isprime(int x){
for (int i = 2; i*i <= x; i++){
if (x%i == 0) return false;
}
return true;
}
int main(int argc, char ** argv) {
int k = 0, i = 2;
ans[0] = 0;
while (k !=... | 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){
int prime[10002];
int sum[10002];
prime[1] = 2;
prime[2] = 3;
sum[1] = 2;
sum[2] = 5;
int ptr=3;
for(int num=5; ptr<=10000; num++){
bool f = false;
for(int i=1; i<ptr; i++){
if(prime[i]*prime[i] > num){
break;
}
if(num%prime[i]==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 <stdio.h>
int s[1000001];
int p[1000001];
int main(void)
{
int i, j, k;
int n;
for (i = 0; i <= 1000000; i++){
s[i] = 1;
}
s[0] = s[1] = 0;
for (i = 2; i * i <= 1000000; i++){
if (s[i] == 0) continue;
for (j = i * i; j <= 1000000; j += i){
s[j] = 0;
}
}
k = 2;
p[1] = 2;
for (i = 3; 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<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#define INF 999999
using namespace std;
bool prime[1000000]={0};
int main(){
prime[1]=1;
for(int i=2;i*i<=1000000;i++){
if(prime[i]==0){
for(int j=i;j<1000000;j+=i){
prime[j]=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 | #include <cmath>
#include <iostream>
#include <vector>
typedef std::vector <int> vi;
vi PRIMES;
vi prime(int n){
vi s,p;
int i,j,x,h;
for (i=1;i<=n;i+=2) s.push_back(i);
h=s.size();
for (i=1;i<=(int)pow(n,0.5);i++)
if (x=s[i]) for (j=i+x;j<h;j+=x) s[j]=0;
s[0]=2;
for (i=0;i<h;i++) if(s[i]!=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 | # coding=utf-8
###
### for atcorder program
###
import sys
import math
# math class
class mymath:
### pi
pi = 3.14159265358979323846264338
### Prime Number
def pnum_eratosthenes(self, n):
ptable = [0 for i in range(n+1)]
plist = []
for i in range(2, n+1):
if ptab... | 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 <algorithm>
#include <vector>
#include <cmath>
using namespace std;
int main(){
int a[120000]={0},b,n,sum,cunt;
for(int i=2;i<120000;i++)
a[i]=1;
b=sqrt(120000);
for(int i=2;i<=b;i++){
if(a[i]==1){
for(int j=2*i;j<120000;j+=i)
a[j]=0;
}
}
while(1){... | CPP |
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