hdu 5150 Sum Sum Sum 水
Sum Sum Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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.
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.
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
3
5 6 7
1
10
Sample Output
12
0
Source
#include<bits/stdc++.h> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-14 const int N=2e5+10,M=1e6+10,inf=1e9+10,mod=1e9+7; const ll INF=1e18+10; int prime(int n) { if(n<=1) return 0; if(n==2) return 1; if(n%2==0) return 0; int k, upperBound=n/2; for(k=3; k<=upperBound; k+=2) { upperBound=n/k; if(n%k==0) return 0; } return 1; } int flag[1010]; int main() { int n,x; for(int i=2;i<=1000;i++) flag[i]=prime(i); flag[1]=1; while(~scanf("%d",&n)) { int ans=0; for(int i=1;i<=n;i++) { scanf("%d",&x); if(flag[x]) ans+=x; } printf("%d\n",ans); } return 0; }