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.
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Print the minimum number of steps modulo 109 + 7.
ab
1
aab
3
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
从后向前 遍历
每个a都会把它后边的b变到后边 且乘二
而每个a后边b的个数 即为当前a变化的次数
#include <bits/stdc++.h> #define mem(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long LL; const int maxn = 1000010, INF = 0x7fffffff; const int MOD = 1e9 + 7; int n; string str; int main() { cin >> str; int len = str.size(); int ans = 0, cnt = 0; for(int i = len - 1; i >= 0; i--) { if(str[i] == 'b') cnt++; else { ans = (ans + cnt) % MOD; cnt = (cnt * 2) % MOD; } } cout << ans << endl; return 0; }
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.
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Print the minimum number of steps modulo 109 + 7.
ab
1
aab
3
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".