AtCoder Beginner Contest 337

A

n=int(input())

a,b=0,0
for i in range(n):
    x,y=map(int,input().split())
    a+=x
    b+=y
if a>b:print("Takahashi")
elif a

B

 A B C必须都连着放，且必须按照ABC的顺序

分别给ABC赋值，有小的出现在大的后边就不符合题意

str=input()

f=0
res=0
for i in str:
    if i=='A':
        x=1
    elif i=='B':
        x=2
    elif i=='C':
        x=3
    if f<=x:
        f=x
    else:
        res=1
        break
if res==0:print("Yes")
else:print("No")
        

C

每两个数都连着一条边，存入字典，依次输出，用前一个数来确定后一个数是什么 

n=int(input())
a=[0]+list(map(int,input().split()))
d={}
idx=0
for i in range(1,n+1):
    if a[i] not in d:
        d[a[i]]=0
    d[a[i]]=i
x=-1
for _ in range(n):
    print(d[x],end=" ")
    x=d[x]
    
        

D

建一个横纵坐标都从1开始的二维数组

统计每一行每一列的前缀和，依次枚举长度为k的数组，且该数组里边没有x

h,w,k=map(int,input().split())

a=[[]]
for i in range(h):
    a.append([0]+list(input()))
zz=max(h,w)
s=[0 for _ in range(zz+1)]
d=[0 for _ in range(zz+1)]
res=1000000000

for i in range(1,h+1):
    for j in range(1,w+1):
        s[j]=s[j-1]+int(a[i][j]=='.')
        d[j]=d[j-1]+int(a[i][j]=='x')
    # print(s)
    # print(d)
    for j in range(k,w+1):
        if d[j]-d[j-k]>0:continue
        else:res=min(res,s[j]-s[j-k])
    # print(res)
for i in range(1,w+1):
    for j in range(1,h+1):
        s[j]=s[j-1]+int(a[j][i]=='.')
        d[j]=d[j-1]+int(a[j][i]=='x')
    for j in range(k,h+1):
        if d[j]-d[j-k]>0:continue
        else:res=min(res,s[j]-s[j-k])
    # print(res)
if res==1000000000:print("-1")
else:print(res)