C. The Fair Nut and String（思维）

版权声明：本文为博主原创文章，转载请说明出处。 https://blog.csdn.net/xianpingping/article/details/84963516

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:

1. 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=
  1. '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

  Input

  Copy

  abbaa
  

  Output

  Copy

  5

  Input

  Copy

  baaaa
  

  Output

  Copy

  4

  Input

  Copy

  agaa
  

  Output

  Copy

  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;
      }
  }