【CodeForces 804B】Minimum number of steps（思维+数学）


[align=center]B. Minimum number of steps[/align]

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".
题目大意：一个只含a,b的字符串，将所有ab替换为bba，求最小步数
思路：从第一个开始替换，最终结果为bbbaaaa这样a在后b在前的字符串，每次替换，多一个b，a的数量不变，并且将a向后移一位，最终目的是将所有a移到b后面，每次操作不影响后面的字符串，可以得出结论对每个ab的操作步数为 ：  设当前b前面所有a
a75d
的个数n，step=2^n - 1；将每个b操作的step加起来就是结果。  这道题有取模运算  (a + b)
% p = (a % p + b % p) % p  ，取模运算写错了233333……ans=(ans + fastpow(2,cot) - 1)%mod；
用strlen会超时，用s.length()不会超时
#include <bits/stdc++.h>
#define manx 100005
typedef long long ll;
const int mod = 1e9+7;
using namespace std;
ll fastpow(ll x,ll n)
{
ll ans=1;
while(n){
if (n&1) ans=(ans*x)%mod;
x=x*x%mod;
n>>=1;
}
return ans;
}
int main()
{
string s;
while(cin >> s){
ll l=s.length(),cot=0,ans=0;
for (int i=0; i<l; i++){
if (s[i]=='a') cot++;
else ans=(ans + fastpow(2,cot) - 1)%mod;
}
printf("%I64d\n",ans);
}
return 0;
}