codeforces 804B——Minimum number of steps（模拟，贪心）-优快云博客

本文链接：https://blog.youkuaiyun.com/say_c_box/article/details/71223889

本文介绍了一个算法问题，即如何将包含字母'a'和'b'的字符串通过一系列操作转换，目标是最小化转换步骤并确保符合特定规则。文章详细解释了转换规则及求解过程，并附带了一个C++实现示例。

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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".

ab->bba,我们可以看到特点就是a的数量不变，并且a都被变到最后一位

从后往前扫，对于每一个a，需要的变换次数是其后的b的个数，而每一次变换，b的个数加一。模拟一下即可

#include<cstdio>
#include<cstring>
#include<queue>
#include <cstdlib>
#include<map>
#include <set>
#include <queue>
#include <iostream>
#include <string>
using namespace std;
const int MAXN = 1000010;
long long INF = 0x7fffffffffffffff;
const long long MODE = 1e9 + 7;
int main() {
	string a;
	cin >> a;
	int n = a.length();
	long long ans = 0;
	long long res = 0;
	for (int i = n - 1; i >= 0; i--) {
		if (a[i] == 'a') {
			res = (res + ans) % MODE;
			ans = (ans*2)%MODE;
		}
		else {
			ans=(ans+1)%MODE;
		}
	}
	cout << res << endl;
	//system("pause");
}