B - Powers of two

Time limit : 2sec / Memory limit : 1024MB

Score : 600 points

Problem Statement

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is Ai. He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2. Note that a ball cannot belong to multiple pairs. Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2t using some non-negative integer t.

Constraints

• 1≤N≤2×105
• 1≤Ai≤109
• Ai is an integer.

Input

Input is given from Standard Input in the following format:

N
A1 A2 … AN

Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Sample Input 1

Copy

3
1 2 3

Sample Output 1

Copy

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls. Note that we cannot pair the second ball with itself.

Sample Input 2

Copy

5
3 11 14 5 13

Sample Output 2

Copy

2

//nqiiii大佬的代码
//看大佬的代码，似乎考虑最优解的情况，即1-15,1-7,7-9，排序后再贪心，但是我看不懂
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, ans;
multiset<ll> st;//因为数据可能重复，所以用multiset
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		ll x; scanf("%lld", &x);
		st.insert(x);
	}
	while (!st.empty()) {
		ll v = *--st.end();//取st最后一位数据
		st.erase(--st.end());//将最后一位数据抹去，避免下面find找到自己
		for (int i = 0; i <= 32; ++i) {//2^32是int范围内的极限,循环int内的2的幂
			ll t = (1ll << i) - v;
			if (st.find(t) != st.end()) {//如果t在st中存在，抹去，ans++，并且break
				++ans; st.erase(st.find(t)); break;
			}
		}
	}
	printf("%d", ans);
}

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