Codeforces805D. Minimum number of steps_we have a string of letters 'a' and 'b'. we want t-CSDN博客

本文链接：https://blog.csdn.net/Jelly_acm/article/details/71190834

本文介绍了一种算法，用于计算将字符串中的所有ab子串替换为bba所需的最小步骤数。通过从后向前遍历字符串并统计b的数量，可以有效地得出答案。

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We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.

The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

Input

The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.

Output

Print the minimum number of steps modulo 109 + 7.

        Examples 
      
          input 
        
ab

          output 
        
1

          input 
        
aab

          output 
        
3

Note

The first example: "ab" → "bba".

The second example: "aab" → "abba" → "bbaba" → "bbbbaa".

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题目的意思是给出一个只含ab的字符串，把里面的ab替换为bba直到没有ab了为止，问最少多少步

思路，最后肯定变成bbb...bbbaaa...aaa的形式，所以不管怎么变步数应该是一样的。分析发现每个a能把他后边的b变成2个b，而a要变的次数一定是后面b的个数，于是从后面开始处理，统计b的数量即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long
const LL mod=1e9+7;

char s[1000006];
int main()
{
    scanf("%s",s);
    int k=strlen(s);
    LL cnt=0;
    LL ans=0;
    int x=0;
    for(int i=k-1; i>=0; i--)
    {
        if(s[i]=='b')
        {
            cnt++;
        }
        else
        {
            ans+=cnt;
            ans%=mod;
            cnt*=2;
            cnt%=mod;
        }

    }
    printf("%lld\n",ans);
    return 0;
}