题目链接:
time limit per test
3 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou 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; }