C. The Fair Nut and String
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
The Fair Nut found a string s
. The string consists of lowercase Latin letters. The Nut is a curious guy, so he wants to find the number of strictly increasing sequences p1,p2,…,pk
, such that:
- For each i
(1≤i≤k), spi=
- 'a'.
- For each i
- (1≤i<k), there is such j that pi<j<pi+1 and sj=
- 'b'.
The Nut is upset because he doesn't know how to find the number. Help him.
This number should be calculated modulo 109+7
.
Input
The first line contains the string s
(1≤|s|≤105) consisting of lowercase Latin letters.
Output
In a single line print the answer to the problem — the number of such sequences p1,p2,…,pk
modulo 109+7.
Examples
InputCopy
abbaa
OutputCopy
5
InputCopy
baaaa
OutputCopy
4
InputCopy
agaa
OutputCopy
3
Note
In the first example, there are 5
possible sequences. [1], [4], [5], [1,4], [1,5].
In the second example, there are 4
possible sequences. [2], [3], [4], [5].
In the third example, there are 3
possible sequences. [1], [3], [4].
-
题意:题意就是选择一段下标递增的子序列,要求满足任意一个都是a,然后中间(不在子序列中的要有一个b)
-
思路:可以这么考虑,假设是abab,这时有【1】,【3】,【1,3】,那么假设是ababa的话就是可以从前边这几个中,再加上新来的a的影响的,多出来的就是:【4】【1,4】【3,4】【1,3,4】,发现就是之前的再多一个。
-
代码:
-
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int mod=1000000000+7; LL sum=0; LL tol=0; LL lala=0; int main() { string s; while(cin>>s){ sum=0; tol=0; int len=s.size(); for(int i=0;i<len;i++){ if(s[i]=='b'){ lala=tol; } if(s[i]=='a'){ tol+=lala+1; tol=tol%mod; } } cout<<tol<<endl; } }