題意：
題解
利用一個數(shù)能被3整除當(dāng)且僅當(dāng)其各位之和sum能被3整除赤炒。
如果sum本身能被3整除氯析，則不需要刪除。
否則統(tǒng)計原數(shù)的每一位數(shù)%3后的個數(shù)莺褒，比較%3 =1與%3 =2 的個數(shù)掩缓，有兩種方法可以使其sum變?yōu)?%3 =0：
%3=1 與%3=2，相互抵消遵岩，還剩下的差值即為答案你辣。
%3=1 與%3=2，先內(nèi)部消化，%3 =1的 三個一消除绢记，%3 =2的三個一消除扁达，最后再相互抵消，差值即為答案蠢熄。
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)
模擬即可跪解。
記錄最遠(yuǎn)位置maxx，當(dāng)前位置res签孔，前綴和sum叉讥，以及前綴和的最大值r。
在每一輪中饥追，首先更新前綴和图仓，然后更新前綴和的最大值，本輪能達(dá)到的最大值顯然是res+r但绕，用其更新maxx救崔，再用res+sum更新res。
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