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  • 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.
     
    Input
    There are several test cases. 
    In each test case:
    The first line contains a integer N(1N1000). 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
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5987419.html
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