Codeforces805D. Minimum number of steps 2017-05-05 08:46 240人阅读 评论(0) 收藏

D. Minimum number of steps

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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".

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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;
}