Code Forces 37A Towers_codeforces 37a-优快云博客

本文链接：https://blog.youkuaiyun.com/lizhaowei213/article/details/47361619

本文介绍了一个利用频率统计方法解决构建木棒塔问题的实例，通过输入不同长度的木棒数量，输出构建出的最大高度和塔的数量。采用高效算法优化，避免重复搜索，实现快速求解。

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B - B

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.

Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.

Input

The first line contains an integer N (1 ≤ N ≤ 1000) — the number of bars at Vasya’s disposal. The second line contains N space-separated integers l_i — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.

Output

In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.

Sample Input

Input

3
1 2 3

Output

1 3

Input

4
6 5 6 7

Output

2 3

水题。

但还是要记录一下，方法很简单，开个数组记录每个元素出现的次数，然后求最大就行了。

当时我没用这种方法，用了一层for循环，试图在循环搜索中，找到最大出现的次数。但后来发现，取最大值的过程写在else里了，没发现的后果就是WA了4次。。。。。。。

之后又用两层for循环的方法实现了一下，复杂度分明。

一层for循环搜索。30ms

#include <stdio.h>
#include <algorithm>
#define N 1005
using namespace std;
int a[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    sort(a,a+n);
    int sum=1;
    int m=1;
    int pos=1;
    for(int i=1;i<n;i++)
    {
        if(a[i]==a[i-1])
        {
            pos++;
        }
        else
        {
            pos=1;
            sum++;
        }
        m=max(m,pos);
    }
    printf("%d %d\n",m,sum);
    return 0;
}

两层for循环搜索 60ms

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 1005
using namespace std;
int a[N];
int pos[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    memset(pos,0,sizeof(pos));
    for(int i=0;i<n;i++)
    {
        if(pos[a[i]]==0)
        {
            for(int j=0;j<n;j++)
            {
                if(a[j]==a[i])
                    pos[a[i]]++;
            }
        }
    }
    int sum=0;
    int m=0;
    for(int i=0;i<1005;i++)
    {
        if(pos[i]!=0)
        {
            sum++;
            m=max(m,pos[i]);
        }
    }
    printf("%d %d\n",m,sum);
    return 0;
}

以此为鉴。