qq_45968329的博客

m = int(input())arr = list(map(int, input().split()))arr = sorted(arr)for i in range(m): print(arr[i], end=' ')

m = int(input())arr = [0 for i in range(m)]for i in range(m): n = input() n = int(n,16) n = "{:o}".format(int(n)) arr[i] = nfor i in range(m): print(arr[i])

m = int(input())m = hex(m)m = m[2:]m = list(m)lg = len(m)for i in range(lg): if int(ord(m[i])) >= 65: n = chr(ord(m[i])-32) m[i] = nfor i in range(lg): print(m[i],end='')

m = int(input())for i in range(10000,1000000): n = str(i) lg = len(n) Sum = 0 if n == n[::-1]: for j in range(lg): Sum += int(n[j]) if Sum == m: print(i)

for i in range(1000,10000): m = str(i) if m == m[::-1]: print(i)

m = int(input())arr = [[0 for i in range(m)] for i in range(m)]if m == 1: print(1)elif m == 2: print(1) print(1,1)else: print(1) print(1,1) arr[1][0] = arr[1][1] = 1 for i in range(2,m): arr[i][0] = arr[i][i] = 1

m = int(input())arr = list(map(int,input().split()))n = int(input())for i in range(m): flog = 0 if arr[i] == n: flog = 1 print(i+1) breakif flog == 0: print(-1)

m = int(input())arr = list(map(int,input().split()))MIN = min(arr)MAX = max(arr)LONG = len(arr)SUM = 0for i in range(LONG): SUM += int(arr[i])print(MAX)print(MIN)print(SUM)

m,n = map(int,input().split())arr = [[0 for i in range(n)] for i in range(m)]for i in range(m): if i <= n: arr[i][0] = chr(ord('A')+i) for j in range(i): arr[i][j] = chr(ord('A')+i-j) for j in range(i,n):

for i in range(32): print("{:05b}".format(int(i)))

注意and和or的使用m = int(input())if m%4==0 and m%100!=0 or m%400==0: print('yes')else: print('no')

m = int(input())if m<=2: print(1)else: fn1 = fn2 = 1 for i in range(2,m): fn = fn1+fn2 fn2 = fn1%10007 fn1 = fn%10007 fn = fn%10007 print(fn)

import math PI = math.pir = int(input())s = pow(r,2)*PIprint('%.7f'%s)

m = int(input())if m%2==0: Sum = int(m/2)*(1+m)else: Sum = int(m/2)*(1+m)+int((1+m)/2)print(Sum)

注意输入格式m,n = map(int, input().split())print(int(m+n))

信道估计Y = HX + N把X理解为要发送的信号，Y是接收信号，那么H就是我们要估计的信道。在发送信号的时候，往往需要发送一些信息已知的导频序列(何为导频？就是先利用已知的导频信息探探路（信道），就像我们平时走路知道知道我们要走的路的状况了，肯定就会容易想到应对措施)。例如发送的导频为X1=b1, X2 =b2， 得到了两个Y1和Y2；就相当于求解方程（大概意思，可能实际并非如此）。再利用已知量求未知数H，也就是估计的信道信息信道均衡 如果理解上面意思，再举个不恰当的例子便于理解。 假如我

​ 先上参考的大佬链接：https://blog.youkuaiyun.com/dog250/article/details/6420427​ 如果看完还不明白，接下来看完我的理解你肯定明白啦！如果有不理解或者错误的请留言讨论，再完善码分复用：​ 首先需要对链接内容看一下，这样就容易理解了。​ 首先，码分复用就是将向各个用户发送的数据叠加成向量L，在接收端重复使用：用户N发送数据时用的沃尔什的K(行或列)，那么就在接收端再用这个沃尔什矩阵的K向量与叠加后的向量相乘求用户数据；而每个用户都要用到这个融合