codeforces Minimum number of steps 思维

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本文介绍了一种算法，用于计算将包含字母'a'和'b'的字符串转换为特定形式所需的最小步骤数。通过记录每个'a'字符后跟随的'b'字符数量并利用这些信息进行计算，最终得出模10^9+7后的结果。

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D. Minimum number of steps
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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”.

简单的思维题保存每个位后面的 b的数量即可。看规律可知，每次遇到一个 a 后面的b的数量会加倍

#include <bits/stdc++.h>
using namespace std;
long long wei[2010100];
const int mod=1e9+7;
int main()
{
    string s;
    cin>>s;
    int len=s.length();
    long long sum=0;
    long long sum1=0;
    for(int i=len-1;i>=0;i--)
    {
        if(s[i]=='b')
        {
            sum++;
            sum%=mod;
        }
        else if(s[i]=='a')
        {
            wei[i]=sum;
            sum+=wei[i];
            sum%=mod;
            wei[i]%=mod;
        }
    }
    long long res=0;
    for(int i=0;i<len;i++)
    {
        res+=wei[i];
        res%=mod;
    }
    printf("%lld\n",res );

}