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".
———————————————————————————————————————
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;
}