CodeForces 702B: Powers of Two（遍历、穷举）

最新推荐文章于 2019-04-18 11:20:20 发布

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#codeforces #csu #acm #power-of-two

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本文介绍了一个算法问题，即找出所有整数对 (i, j)，使得 a[i] + a[j] 是2的幂次。提供了两种解决方案：遍历查找法和使用哈希映射优化时间复杂度的方法。

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Powers of Two
time limit per test: 3 seconds
memory limit per test: 256 megabytes
inputstandard input
outputstandard output
You are given n integers

a1, a2, ..., an $a_1, a_2, ..., a_n$ . Find the number of pairs of indexes i, j (i < j) that

ai + aj $a_i + a_j$ is a power of 2 (i. e. some integer x exists so that

ai + aj = 2x $a_i + a_j = 2^x$ ).
Input

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

The second line contains n positive integers $a_1, a_2, ..., a_n$ (1 ≤ $a_i$ ≤ 109).

Output

Print the number of pairs of indexes i, j (i < j) that $a_i + a_j$ is a power of 2.

Simple Input

Simple Output

2
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.

Code 1
思路：遍历查找

//时间复杂度太高，当输入数据组数较多时会超时
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long LL;
#define maxn 100005
LL a[maxn]={0};

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int sum=0;
        for(int i=0; i<n; i++)
            scanf("%lld",&a[i]);
        for(int i=0; i<n-1; i++)
            for(int j=i+1; j<n; j++){
                LL x=1;
                for(int k=0; k<=31; k++)
                    {
                        if(a[i]+a[j]==x) {sum++;break;}
                        x=x<<1;
                    }
            }
        printf("%d\n",sum);
    }
    return 0;
}

Code 2
思路：利用 $a_i+a_j=2^x$ 关系枚举

#include<bits/stdc++.h>
using namespace std;
#define LL long long
map<LL,int>a;           //映射到int的初始值是零
LL sum=0;

int main()
{
    int n;
    while(~scanf("%d",&n)){
    for(int i=0; i<n; i++)
    {
        LL x;
        scanf("%lld",&x);
        for(int j=0; j<=31; j++)
            sum+=a[(1LL<<j)-x];     //由ai+aj=2^x枚举x,若对应LL非空，则加上其映射值int
        a[x]++;                     //用int记录对应LL出现的次数
    }
    printf("%lld\n",sum);
    }
    return 0;
}