输出次数HDU2192:MagicBuilding

新手发帖，很多方面都是刚入门，有错误的地方请大家见谅，欢迎批评指正

    

Problem Description

    

As the increase of population, the living space for people is becoming smaller and smaller. In MagicStar the problem is much worse. Dr. Mathematica is trying to save land by clustering buildings and then we call the set of buildings MagicBuilding. Now we can treat the buildings as a square of size d, and the height doesn't matter. Buildings of d1,d2,d3....dn can be clustered into one MagicBuilding if they satisfy di != dj(i != j).  
Given a series of buildings size , you need to calculate the minimal numbers of MagicBuildings that can be made. Note that one building can also be considered as a MagicBuilding.
Suppose there are five buildings : 1, 2, 2, 3, 3. We make three MagicBuildings (1,3), (2,3), (2) .And we can also make two MagicBuilding :(1,2,3), (2,3). There is at least two MagicBuildings obviously.

    

 

    

Input

    

The first line of the input is a single number t, indicating the number of test cases.
Each test case starts by n (1≤n≤10^4) in a line indicating the number of buildings. Next n positive numbers (less than 2^31) will be the size of the buildings.

    

 

    

Output

    每日一道理
记不清有多少个夜晚，在我翻阅纸张的指间滑落；记不清有多少支蜡烛，在我的凝视中化为灰烬。逝者如斯，我时时刻刻会听见自己对生命承诺的余音，感到岁月的流转在渐渐稀释我的年少无知，我愿自己是一只上足了发条的时钟，在昼夜不停的流转中留下自己充实的每一刻。

    

For each test case , output a number perline, meaning the minimal number of the MagicBuilding that can be made.

    

 

    

Sample Input

2 1 2 5 1 2 2 3 3

 

    

Sample Output

1 2

 

    
直接输出出现最多的次数便可

#include <stdio.h>
#include <algorithm>
using namespace std;

int a[1000000];

int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int m,i;
        scanf("%d",&m);
        for(i = 0;i<m;i++)
        scanf("%d",&a[i]);
        sort(a,a+m);
        int l = 1,tem = a[0],max = 1;
        for(i = 1;i<m;i++)
        {
            if(a[i] == tem)
            {
                l++;
            }
            else
            {
                tem = a[i];
                l = 1;
            }
            if(l > max)
            max = l;
        }
        printf("%d\n",max);
    }

    return 0;
}

    
 

文章结束给大家分享下程序员的一些笑话语录： 不会，Intel会维持高利润，也会维持竞争局面，国外的竞争不是打死对方的那种。你看日本有尼康，佳能，索尼，都做相机，大家都过得很滋润。别看一堆厂，其实真正控制的是后面的那几个财团——有些竞争对手，后面其实是一家人。

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