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
Note
In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).
In the second example all pairs of indexes (i, j) (where i < j) include in answer.
n个数字，问有多少对数字加起来刚好是2的k次方。
这还用说？枚举个k再枚举个a[i]然后看看有没有2^k-a[i]这个数就好了
这里我为了防被x没用hash用了二分
不过要考虑一个数字出现很多次的情况，或者你要找的刚好就是这个数的情况
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<set> 8 #include<map> 9 #include<ctime> 10 #define LL long long 11 #define inf 0x7ffffff 12 #define pa pair<int,int> 13 #define pi 3.1415926535897932384626433832795028841971 14 using namespace std; 15 inline LL read() 16 { 17 LL x=0,f=1;char ch=getchar(); 18 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 19 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 20 return x*f; 21 } 22 inline void write(LL a) 23 { 24 if (a<0){printf("-");a=-a;} 25 if (a>=10)write(a/10); 26 putchar(a%10+'0'); 27 } 28 inline void writeln(LL a){write(a);printf(" ");} 29 int n,sav; 30 LL ans; 31 int a[100010]; 32 int rep[100010]; 33 int bin[110]; 34 inline bool bsearch(int l,int r,int x,int dat) 35 { 36 if (dat<=0)return 0; 37 if (l>r)return 0; 38 int ans=-1; 39 while (l<=r) 40 { 41 int mid=(l+r)>>1; 42 if (a[mid]==dat){ans=mid;break;} 43 if (a[mid]>dat)r=mid-1; 44 if (a[mid]<dat)l=mid+1; 45 } 46 sav=ans; 47 return (ans!=x||ans==x&&rep[x]>1)&&a[ans]==dat; 48 } 49 int main() 50 { 51 n=read(); 52 bin[0]=1; 53 for (int i=1;i<31;i++)bin[i]=bin[i-1]*2; 54 for(int i=1;i<=n;i++)a[i]=read(); 55 sort(a+1,a+n+1); 56 int cur=0; 57 for(int i=1;i<=n;i++) 58 { 59 if (a[i]!=a[i-1])a[++cur]=a[i],rep[cur]=1; 60 else rep[cur]++; 61 } 62 n=cur; 63 for(int i=1;i<=n;i++) 64 { 65 for (int j=0;j<31;j++) 66 if (bsearch(i,n,i,bin[j]-a[i])) 67 { 68 if (sav==i)ans+=(LL)rep[i]*(rep[i]-1)/2; 69 else ans+=(LL)rep[i]*rep[sav]; 70 } 71 } 72 printf("%lld ",ans); 73 }