【Codeforces】-702B-Powers of Two(二分)
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B. Powers of Two
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
47 3 2 1
output
2
input
31 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.
题解:有一种位运算判断是不是2^n 的方法,但是在这里还是会超时,这里用二分。
方法如图,啊偶,怎么把和师父的聊天记录贴出来了?我什么都布吉岛~~~
,这个方法可以了解,感觉不错的
下面是二分的代码:
<span style="font-size:10px;">#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))int main(){int n;__int64 a[100010];while(~scanf("%d",&n)){CLR(a,0);__int64 num=0;//这里num用__int64不然会wa for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);sort(a+1,a+1+n);//排序 for(int i=1;i<=n;i++){for(int j=1;j<=31;j++){__int64 t=((__int64)1<<j)-a[i];//t=2^n-a[i]再在剩余的数里找有没有是t的数 num+=upper_bound(a+1+i,a+1+n,t)-lower_bound(a+1+i,a+1+n,t);}//大于t的开始的点的地址-大于等于t开始的点的地址,他们相差的数目,就是t的个数 }printf("%I64d\n",num);}return 0;}</span>
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