【思维】ABC

题目描述

You are given a string s consisting of A, B and C.
Snuke wants to perform the following operation on s as many times as possible:
Choose a contiguous substring of s that reads ABC and replace it with BCA.
Find the maximum possible number of operations.
Constraints
1≤|s|≤200000
Each character of s is A, B and C.

输入

Input is given from Standard Input in the following format:
S

输出

Find the maximum possible number of operations.

样例输入

ABCABC

样例输出

3

提示

You can perform the operations three times as follows: ABCABC → BCAABC → BCABCA → BCBCAA. This is the maximum result.

 

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参考：SZG大佬的题解

题意

在一个只包含A、B、C的字符串，有一种操作，可使 “ABC” 变成 ”BCA“，求字符串s的最多操作数。

1≤∣s∣≤200000

思路

  易得，该操作是将A与BC交换位置，可用 1、0分别代表“A”、“BC”。题意转化对一个只包含10的序列，

将所有的10更新01，即将所有的0放在1前面。假设序列中共有kk个0，每个0前面有ai​个1，则ans=∑​ai （1,k）​

  对于单独B、C，则可看作是两个序列分隔的标志。

 

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#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+100;
typedef long long ll;
char s[N];
int main()
{
    scanf("%s",s);
    ll cnt = 0 ;
    ll ans = 0 , i = 0 ;
    int len = strlen(s);
    while ( s[i] ) {
        if( s[i] == 'A' ){
            //printf("#1 %d \n" ,i);
            cnt ++ ;
            i++ ;
        }else if ( s[i] == 'B' && s[i+1] == 'C' ){
            //printf("#2 %d \n" ,i);
            ans = ans + cnt ;
            i+=2 ;
            if( i >= len ) break ; 
        }else if ( s[i] == 'B' || s[i] == 'C' || s[i] == '\0' ){
            cnt = 0 ;
            i++ ;
        }
    }
    //ans = ans + cnt ;
    printf("%lld\n",ans);
}