2015 HUAS Summer Trainning #5~E

最新推荐文章于 2021-07-24 09:11:52 发布

转载最新推荐文章于 2021-07-24 09:11:52 发布 · 129 阅读

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原文链接：http://www.cnblogs.com/chenchunhui/p/4722462.html

本文介绍了一种高效算法，用于在股票价格序列中快速找到最长上升趋势。通过输入股票价格序列的长度和具体数值，输出最长上升趋势的长度。解决实际投资决策中的关键问题。

Description

The world financial crisis is quite a subject. Some people are more relaxed while others are quite anxious. John is one of them. He is very concerned about the evolution of the stock exchange. He follows stock prices every day looking for rising trends. Given a sequence of numbers p1, p2,...,pn representing stock prices, a rising trend is a subsequence pi1 < pi2 < ... < pik, with i1 < i2 < ... < ik. John’s problem is to find very quickly the longest rising trend.

Input

Each data set in the file stands for a particular set of stock prices. A data set starts with the length L (L ≤ 100000) of the sequence of numbers, followed by the numbers (a number fits a long integer).
White spaces can occur freely in the input. The input data are correct and terminate with an end of file.

Output

The program prints the length of the longest rising trend.
For each set of data the program prints the result to the standard output from the beginning of a line.

Sample Input

6 
5 2 1 4 5 3 
3  
1 1 1 
4 
4 3 2 1

Sample Output

3 
1 
1

Hint

There are three data sets. In the first case, the length L of the sequence is 6. The sequence is 5, 2, 1, 4, 5, 3. The result for the data set is the length of the longest rising trend: 3.

解题思路：这个题目要小心时间问题，稍不小心就会超时。还要注意的是，题目中说的最大升序并不一定是要连续的！！这是非常要注意的问题。

程序代码：

#include <iostream>
#include<stdio.h>
using namespace std;
int a[150000],b[150000],t;
void dog(int z)
{
    int x,y,mid;
    x=1;
    y=t;
    while(x<=y)
    {
        mid=(x+y)/2;
        if (b[mid]==z) return ;
        else
            if (b[mid]>z) y=mid-1;
            else
                x=mid+1;
    }
    if (x==t+1)
    {
        t++;
        b[t]=z;
    }
    else
        b[x]=z;
}
int main() {
    int n,i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        t=1;
        b[1]=a[1];
        for(i=2;i<=n;i++)
        {
            dog(a[i]);
        }
        printf("%d\n",t);
    }
    return 0;
}

转载于:https://www.cnblogs.com/chenchunhui/p/4722462.html