注意问题：重复统计；数组个数输入确定；计数方法；计算量少
#include<iostream>
using namespace std;int i_zhaungtai=0;//状态量，判断
int number,x;//number记录输入范围，x记录输入的值
int a[100];//输入数组个数记录
int main()
{
cout << "请键入所输值的范围：" << '\n';
cin >>number;
//count c = i_x;
int *i_s =new int[number];//将键入的number给数组大小
for (int i = 0; i < number;++i)
i_s[i]=0;//数组初始化
int i_tem;//记录输入值
while (cin >> i_tem)
{
i_s[i_tem] = i_s[i_tem] + 1;//计数式
if (i_s[i_tem] == 1)//判断出现次数超过一就不写入，保证a中都是唯一值
{
a[x] = i_tem;
x = x + 1;
}
}
/*for (int i = 0; i < x; ++i)
{
cout << a[i] << '\n';
}*/
/*cin >> i_tem;
cout << i_tem << '\n';
int a[5] = { 1, 2, 3, 0, 6 };
for (int i = 0; i < 5; ++i)
i_s[a[i]] = i_s[a[i]] + 1;*/
for (int i = 0; i <x; ++i)
{
if (i_s[a[i]]>1)//只判断输入个数的值，不必全部范围内检测
{
cout << a[i] << "重复出现" << i_s[a[i]] << "次" << '\n';
i_zhaungtai = 1;
}
}
if (i_zhaungtai==0)//判断状态量
cout << "无重复数字" << '\n';
/*cout << '\n';
cout << x << ' '<<number<<'\n';
for (int i = 0; i < number; ++i)
{
cout << i_s[i] << '\n';
}*/
system("pause");
}
import numpy as np a=(raw_input()).split() array1=np.array(a) #print(array1.dtype) #print(array1) uniqs, counts = np.unique(array1, return_counts=True) #print("Unique items : ", uniqs) #print("Counts : ", counts) for i in uniqs: if counts[np.where(uniqs==i)]>1: print (i,'has appeared',int(counts[np.where(uniqs==i)]),'times')
#encoding =utf-8 import numpy as np #####date is producted by keybord #a=(raw_input()).split() #array1=np.array(a) #####date is producted by random rn = np.random.RandomState(100) array1=rn.randint(0,10,size=[1,20]) print(array1) #print(array1.dtype) #print(array1) uniqs, counts = np.unique(array1, return_counts=True) print("Unique items : ", uniqs) print("Counts : ", counts) #####using for make the classification #for i in range(0,len(uniqs)): # if counts[i]>1: # print (uniqs[i],'has appeared',int(counts[i]),'times') #####using np.where make the classification index1=np.where(counts>2) print(uniqs.take(index1),"has appeared more than 2 times")
如果对你有帮助，谢谢你^_^
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