问题描述

给定n个整数，请统计出每个整数出现的次数，按出现次数从多到少的顺序输出。

输入格式

输入的第一行包含一个整数n，表示给定数字的个数。

第二行包含n个整数，相邻的整数之间用一个空格分隔，表示所给定的整数。

输出格式

输出多行，每行包含两个整数，分别表示一个给定的整数和它出现的次数。按出现次数递减的顺序输出。如果两个整数出现的次数一样多，则先输出值较小的，然后输出值较大的。

样例输入

12
5 2 3 3 1 3 4 2 5 2 3 5

样例输出

评测用例规模与约定

1 ≤ n ≤ 1000，给出的数都是不超过1000的非负整数。

编程语言：Python

解题思想：

1.首先重开一个列表来存放去除重复后的初始列表。

2.然后判断计数，按照题目要求进行排序并按格式输出。

时间复杂度：O(n^2)

IDE代码如下：

N = int(input())  # 数字的个数
Number = [0 for i in range(N)]  # 定义存放数字的列表
Number = [int(n) for n in input().split()]  # 将一行的值存入列表
NewNumber = list(set(Number))  # 去除重复后在转为列表存放到另外一个列表中
NewNumberCount = [NewNumber, [0 for i in range(len(NewNumber))]]  # 存放去除重复后每个元素的值和计数
# 例如 NewNumber =[1,2,3,4,5] 则NewNumberCount = [[1,2,3,4,5],[0,0,0,0,0]]
 
for i in range(N):  # 遍历Number列表
    # 无需进行任何判断，因为Number列表中的每个元素都必定存在于NewNumber和NewNumberCount中
    NewNumberCount[1][NewNumber.index(Number[i])] += 1
    # NewNumber.index(Number[i])用来获得Number[i]在NewNumber中的下标
    # NewNumberCount[1][NewNumber.index(Number[i])]用来获得对应的计数
 
 
# 开始按照题目要求进行冒泡排序
def Sort():  # 元素排序
    swap = NewNumberCount[1][j]
    NewNumberCount[1][j] = NewNumberCount[1][i]
    NewNumberCount[1][i] = swap
 
def CountSort():  # 计数排序
    swap = NewNumberCount[0][j]
    NewNumberCount[0][j] = NewNumberCount[0][i]
    NewNumberCount[0][i] = swap
 
 
for i in range(len(NewNumber)):
    for j in range(i + 1, len(NewNumber)):
        if NewNumberCount[1][j] > NewNumberCount[1][i]:
            Sort()
            CountSort()
        elif NewNumberCount[1][j] == NewNumberCount[1][i]:
            if NewNumberCount[0][j] < NewNumberCount[0][i]:
                Sort()
                CountSort()
 
#按照题目格式输出
for i in range(len(NewNumber)):
    print(str(NewNumberCount[0][i])+" "+str(NewNumberCount[1][i]))

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CCF-CSP_20150302:数字排序