题意：
题解
利用一个数能被3整除当且仅当其各位之和sum能被3整除。
如果sum本身能被3整除，则不需要删除。
否则统计原数的每一位数%3后的个数，比较%3 =1与%3 =2 的个数，有两种方法可以使其sum变为 %3 =0：
%3=1 与%3=2，相互抵消，还剩下的差值即为答案。
%3=1 与%3=2，先内部消化，%3 =1的 三个一消除，%3 =2的三个一消除，最后再相互抵消，差值即为答案。
IA=lambda:map(int,input().strip().split())s=input()n=len(s)num=[0foriinrange(3)]summ=0foritins:x=int(it)num[x%3]+=1summ=(summ+x)%3ifsumm%3==0:print(0)else:# print(num)cha=abs(num[1]-num[2])num[1]%=3num[2]%=3cha=min(cha,abs(num[1]-num[2]))ifcha==n:print(-1)else:print(cha)
IA = lambda: map(int, input().strip().split())
s = input()
n = len(s)
num = [0 for i in range(3)]
summ = 0
for it in s:
x = int(it)
num[x % 3] += 1
summ = (summ + x) % 3
if summ % 3 == 0:
print(0)
else:
# print(num)
cha = abs(num[1] - num[2])
num[1] %= 3
num[2] %= 3
cha = min(cha, abs(num[1] - num[2]))
if cha == n:
print(-1)
else:
print(cha)
模拟即可。
记录最远位置maxx，当前位置res，前缀和sum，以及前缀和的最大值r。
在每一轮中，首先更新前缀和，然后更新前缀和的最大值，本轮能达到的最大值显然是res+r，用其更新maxx，再用res+sum更新res。
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