cf 805d

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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’ so，mewhere 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”.

大意 给你一个只有a，b组成的字符串你可以对他进行操作每次将一个ab换成bba当字符串中不存在ab操作就终止问你最小的操作次数。
可以看出每将一个a移动到后面他所需要的次数就是a后面b的个数与此同时b也将翻倍。

#include<bits/stdc++.h>
using namespace std;
long long p=1e9+7;
int main()
{int sum=0,ans=0;char a[1000009];while(~scanf("%s",a)){sum=0,ans=0;int l=strlen(a);for(int i=l-1;i>=0;i--){if(a[i]=='a'){sum+=ans;ans*=2;ans%=p;sum%=p;}elseans++;}sum%=p;cout<<sum<<endl;}
}

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