HDOJ 5150 Sum Sum Sum (素数和 水)

Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 982    Accepted Submission(s): 578

Problem Description

We call a positive number X P-number if there is not a positive number that is less than X and the greatest common divisor of these two numbers is bigger than 1.
Now you are given a sequence of integers. You task is to calculate the sum of P-numbers of the sequence.

 

Input

There are several test cases.
In each test case:
The first line contains a integer N(1≤N≤1000). The second line contains N integers. Each integer is between 1 and 1000.

 

Output

For each test case, output the sum of P-numbers of the sequence.

 

Sample Input

35 6 7110

 

Sample Output

120

 

题意：n个数，求素数和

思路：水题，直接打表，注意这里1是素数



ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1.0)using namespace std;int gcd(int a,int b){return b?gcd(b,a%b):a;}LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}//headint v[MAXN];void db(){mem(v);for(int i=2;i<=1000;i++){if(!v[i]){for(int j=i*2;j<=1000;j+=i)v[j]=1;}}}int main(){db();int n;while(scanf("%d",&n)!=EOF){int ans=0;for(int i=0;i<n;i++){int a;scanf("%d",&a);if(!v[a])ans+=a;}printf("%d\n",ans);}return 0;}