Codeforces805D. Minimum number of steps

最新推荐文章于 2021-04-10 20:05:45 发布

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本文介绍了一个算法问题，针对由字符 'a' 和 'b' 组成的字符串，通过将子串 'ab' 替换为 'bba' 的方式，求解达到最终状态所需的最少操作次数。文章提供了一种从右向左遍历字符串的方法，并详细解释了计算步骤。

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.

题目大意：给定只有ab的字符串遇到ab就变成bba 问字符串需要改变多少次

思路：从右往左走 "aab" 遇到a b的数量翻倍答案加1(b的数量) "abba" 又遇到a b的数量又翻倍（答案加2 因为遇到a之前有两个b） → "bbbbaa".

#include <stdio.h>
#include <bits/stdc++.h>
#define mod 1000000007
typedef long long ll;
using namespace std;
char s[1000005];
int main()
{
	 scanf("%s",s);
	 int len=strlen(s)-1;
	 ll b=0,ans=0;
	 for(int i=len;i>=0;i--){
	 	if(s[i]=='a'){
	 		ans=(ans+b)%mod;
	 		b=b*2%mod;
	 	}else{
	 		b++;b%=mod;
	 	}
	 }
	 cout<<ans<<endl;
	 return 0;
 }