Codeforces Round #411 (Div. 2) D. Minimum number of steps

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".

题意 ：

ab串必须全部转化位bba

我们需要的操作就是统计每个b前边a的数量

已知 ab需要一次

aab需要三次 1+2

aaab需要 7次 1+2+4(2^0+2^1+2^2==2^num-1);

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int N=1e9+7; #define ll long long int dp[1000333]; ll quick_mi(int n) { ll res=1; ll tem=2; while(n) { if(n%2==1) res*=tem; tem*=tem; tem%=N; res%=N; n/=2; } return res; } char s[1000333]; int main() { scanf("%s",s); int l=strlen(s); if(s[0]=='a') dp[0]=1; else dp[0]=0; long long res=0; for(int i=1;i<l;i++) { if(s[i]=='a') dp[i]=dp[i-1]+1; else { dp[i]=dp[i-1]; res+=(quick_mi(dp[i])-1); res%=N; } } printf("%lld\n",res); }