• codeforces 702B B. Powers of Two(水题)


    题目链接:

    B. Powers of Two

    time limit per test
    3 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer xexists so that ai + aj = 2x).

    Input

    The first line contains the single positive integer n (1 ≤ n ≤ 105) — the number of integers.

    The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

    Output

    Print the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2.

    Examples
    input
    4
    7 3 2 1
    output
    2
    input
    3
    1 1 1
    output
    3

    题意:
    问有多少对a[i]+a[j]是2的次幂;

    思路:

    map搞一搞;

    AC代码:

    /************************************************ 
    ┆  ┏┓   ┏┓ ┆    
    ┆┏┛┻━━━┛┻┓ ┆ 
    ┆┃       ┃ ┆ 
    ┆┃   ━   ┃ ┆ 
    ┆┃ ┳┛ ┗┳ ┃ ┆ 
    ┆┃       ┃ ┆  
    ┆┃   ┻   ┃ ┆ 
    ┆┗━┓    ┏━┛ ┆ 
    ┆  ┃    ┃  ┆       
    ┆  ┃    ┗━━━┓ ┆ 
    ┆  ┃  AC代马   ┣┓┆ 
    ┆  ┃           ┏┛┆ 
    ┆  ┗┓┓┏━┳┓┏┛ ┆ 
    ┆   ┃┫┫ ┃┫┫ ┆ 
    ┆   ┗┻┛ ┗┻┛ ┆       
    ************************************************ */  
    
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    
    using namespace std;
    
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
    
    typedef  long long LL;
    
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
    
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const int inf=1e9;
    const int N=1e5+10;
    const int maxn=(1<<8);
    const double eps=1e-8;
    
    int a[N],cnt=0,b[50];
    
    map<int,int>mp,vis;
    
    inline void Init()
    {
        LL temp=2;
        while(temp<=2e9+100)
        {
            b[++cnt]=(int)temp;
            temp=temp*2;
        }
    }
    int main()
    {       Init();
            int n;
            read(n);
            For(i,1,n)read(a[i]),mp[a[i]]++,vis[a[i]]=1;
            LL ans=0;
            For(i,1,n)
            {
                For(j,1,cnt)
                {
                    if(vis[b[j]-a[i]])
                    {
                        int x=b[j]-a[i];
                        if(x==a[i])
                        {
                            ans=ans+mp[x]-1;
                        }
                        else 
                        {
                            ans=ans+mp[x];
                        }
                    }
                }
            }
            cout<<ans/2<<endl;
            
            return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5720273.html
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