Codeforces Round #411 D. Minimum number of steps （贪心。）

最新推荐文章于 2025-08-01 11:12:48 发布

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最新推荐文章于 2025-08-01 11:12:48 发布

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本文介绍了一个字符串转换问题的解决方法，对于由字符'a'和'b'组成的字符串，通过将所有出现的子串“ab”替换为“bba”的操作，求达到所有可能替换操作完成所需的最小步骤数。

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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

output

input

aab

output

Note

The first example: “ab” → “bba”.

The second example: “aab” → “abba” → “bbaba” → “bbbbaa”.

参考http://blog.youkuaiyun.com/dormousenone/article/details/71190792
**#解题思路
根据题意，转换规则如下：
ab→bba
abb→bbbba
abbb→bbbbbba**

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
char s[1000010];
int main()
{
    scanf(" %s",s);
    int len = strlen(s);
    long long cntb = 0, ans = 0;
    for(int i=len-1;i>=0;--i)
    {
        if(s[i] == 'a') {
            (ans += cntb) %= mod;
            (cntb *= 2) %= mod;
        } else {
            cntb++;
        }
    }
    printf("%I64d\n", ans%mod);
}