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AI处理器组合--华为OD机试2024年E卷

news 2025/2/5 18:40:55

题解：

代码：

def AI_clq(ll,num):l1=[i for i in ll if i<4]l2=[i for i in ll if i>=4]ans=[]if num==8:if len(l1)==4 and len(l2)==4:ans.append(l1+l2)elif num==4:if len(l1)==4:ans.append(l1)if len(l2)==4:ans.append(l2)elif num==2:if len(l1)==2 or len(l2)==2:if len(l1)==2:ans.append(l1)if len(l2)==2:ans.append(l2)elif len(l1)==4 or len(l2)==4:if len(l1)==4:ans.append([l1[0], l1[1]])ans.append([l1[0], l1[2]])ans.append([l1[0], l1[3]])ans.append([l1[1], l1[2]])ans.append([l1[1], l1[3]])ans.append([l1[2], l1[3]])if len(l2)==4:ans.append([l2[0], l2[1]])ans.append([l2[0], l2[2]])ans.append([l2[0], l2[3]])ans.append([l2[1], l2[2]])ans.append([l2[1], l2[3]])ans.append([l2[2], l2[3]])else:if len(l1)==3:ans.append([l1[0], l1[1]])ans.append([l1[0], l1[2]])ans.append([l1[1], l1[2]])if len(l2)==3:ans.append([l2[0], l2[1]])ans.append([l2[0], l2[2]])ans.append([l2[1], l2[2]])else:if len(l1)==1 or len(l2)==1:if len(l1)==1:ans.append(l1)if len(l2)==1:ans.append(l2)elif len(l1)==3 or len(l2)==3:  #剩余3个处理器优先if len(l1)==3:ans.append([l1[0]])ans.append([l1[1]])ans.append([l1[2]])if len(l2)==3:ans.append([l2[0]])ans.append([l2[1]])ans.append([l2[2]])elif len(l1)==2 or len(l2)==2:   #剩余2个处理器优先if len(l1)==2:ans.append([l1[0]])ans.append([l1[1]])if len(l2)==2:ans.append([l2[0]])ans.append([l2[1]])else:                             #剩余4个处理器if len(l1)==4:ans.append([l1[0]])ans.append([l1[1]])ans.append([l1[2]])ans.append([l1[3]])if len(l2)==4:ans.append([l2[0]])ans.append([l2[1]])ans.append([l2[2]])ans.append([l2[3]])return ansdef arry_str(ans):n=", ".join(["["+", ".join(map(str,v))+"]" for v in ans])return "["+n+"]"ll=[0,1,2,3,4,5,6,7]
num=4
ans=AI_clq(ll,num)      #结果
res=arry_str(ans)       #将结果转换为牛客所需输出格式
print(ans)
print(res)

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