美团比赛

 

'''
第一题:
'''
num=int(input())


#设tmp是要处理的字符串
#如何按方向键最少,感觉就只有一种方案啊.
#先改编码
dict={'ABC':(1,2),'DEF':(1,3),'GHI':(2,1),'JKL':(2,2),'MNO':(2,3),
      'PQRS':(3,1),'TUV':(3,2),'WXYZ':(3,3)}
last=(1,1)
sum=0
for i in range(num):
    tmp=input()
    last=(1,1)
    sum=0
    for jj in tmp:
        for j in dict:
            if jj in j:
                now=dict[j]
                break
        #now是jj的编码
        sum+=abs(now[0]-last[0])+abs(now[1]-last[1])
        last=now

'''
第二题:棋盘
'''
#数组b的最右面少的需要数组a最左边的搬过来
num=int(input())
list1=input().split(' ')
list2=input().split(' ')
for i in range(len(list1)):

    list1[i]=int(list1[i])
for i in range(len(list2)):
    list2[i]=int(list2[i])
sum=0
tmp=0
i=len(list1)-1
while  i in range(len(list1)-1,-1,-1):
    if list1[i]<list2[i]:
        que=list2[i]-list1[i]
        #找到最前面是1的那个位
        
        while 1:
            if list1[tmp]==0:
                tmp+=1
                continue
            else:
                if list1[tmp]>=que:
                    list1[tmp]-=que
                    list1[i]+=que
                    sum+=que*(tmp+i)
                    break
                else:
                    
                    list1[i]+=list1[tmp]
                    sum+=list1[tmp]*(tmp+i)
                    list1[tmp]=0
                    break
       
    
    if list1[i]==list2[i]:
        i-=1
        continue
    if list1[i]>list2[i]:
        sum+=list1[i]-list2[i]
        list1[i-1]+=-(list2[i]-list1[i])
        list1[i]-=-(list2[i]-list1[i])
        
        i-=1
print(sum)

View Code

 

    print(sum)

 

转载于:https://www.cnblogs.com/zhangbo2008/p/9160592.html