D. Minimum number of steps time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard 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”.
简单的思维题 保存每个位后面的 b的数量即可。看规律可知，每次遇到一个 a 后面的b的数量会加倍
#include <bits/stdc++.h> using namespace std; long long wei[2010100]; const int mod=1e9+7; int main() { string s; cin>>s; int len=s.length(); long long sum=0; long long sum1=0; for(int i=len-1;i>=0;i--) { if(s[i]=='b') { sum++; sum%=mod; } else if(s[i]=='a') { wei[i]=sum; sum+=wei[i]; sum%=mod; wei[i]%=mod; } } long long res=0; for(int i=0;i<len;i++) { res+=wei[i]; res%=mod; } printf("%lld\n",res ); }