Little Sub and Counting

最新推荐文章于 2023-02-20 12:29:49 发布

南巷清风i

最新推荐文章于 2023-02-20 12:29:49 发布

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本文链接：https://blog.youkuaiyun.com/YT2017WQJ/article/details/89840913

博客提出一个正整数序列计数问题，给定正整数序列A，需计算对于每个Ai，序列中满足AiAj > AjAi的元素Aj的数量。还给出了输入输出要求，以及lower_bound函数的注释说明。

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题目描述

Little Sub likes Math. Now he has a simple counting problem for you.
Given an positive integer sequence A, for each Ai , Please calculate how many elements Aj in the sequence satisfi ed AiAj > AjAi .

输入

The first line contains one positive integer n(1 ≤ n≤100000).
The following line contains n positive integers Ai(1≤Ai≤231-1).

输出

Please output n integer in one line and separate them by a single space. The ith integer indicates the answer concerning Ai .

样例输入

复制样例数据

3
1 2 5

样例输出

0 2 1

注释：①lower_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于或等于num的数字，格式：lower_bound(a,a+n,key)-a;

②

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
double a[100010],b[100010],x;//注意用double，避免精度问题
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lf",&x);
        a[i]=log(x)/x;
        b[i]=a[i];
    }
    sort(b,b+n);
    int m;
    for(int i=0;i<n;i++){
        m=lower_bound(b,b+n,a[i])-b;
        if(i==n-1)
            printf("%d\n",m);
        else
            printf("%d ",m);
    }
    return 0;
}