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”.
题意：有一个只含有a和b的串，要让b全部到左边，并且移动的规则为 “ab”->”bba”。 分析： 只需要记录每个b前面有多少个a，因为b移动一个a为2^0,移动两个a为 2^0+2^1，所以求等比数列和就可以了。
我用了比较无脑的写法，把所有要的都处理出来然后直接计算~~
AC代码：
#include<stdio.h> #include<string.h> #define mod 1000000007 char a[2000000]; long long b[2000000]; long long c[2000000]; long long d[2000000]; int main() { d[1]=1; c[1]=1; for(int i=2;i<=1000000;i++) { d[i]=(d[i-1]*2)%mod; c[i]=(c[i-1]+d[i])%mod; } scanf("%s",a); int count=0,t=0; for(int i=0;i<strlen(a);i++) { if(a[i]=='a') count++; else b[t++]=count; } long long ans=0; for(int i=0;i<t;i++) { ans+=c[b[i]]; ans%=1000000007; } printf("%lld\n",ans); }