Codeforces-1084C

---

title: Codeforces-1084C

date: 2018-12-13 16:02:04

tags:

• acm
• 刷题
  
  categories:
• Codeforces

---

概述

好久没写博客了，，，最近的事太多了，，几乎没有专门看一个知识点，，，一直在做各种的简单题QAQ。。。

题意

这道题大概的题意就是给你一个串，，，然后找出所有开头结尾都是a的并且中间有b的子串的个数，，，单一的a也算，，，

一开始看了好几遍都没看懂题意，，，不知道在求啥，，，

然后看了一个人的题解，，然后没看懂QAQ,,,,,

看了官方题解下的一个评论看懂了，，，，，emmmm

It doesn't depend at all if there exists any letter other than a or b in the given string. You can for sure ignore those letters, so the editorial says to erase them. Now, what you have is a string consisting only of a and b's. Also two consecutive b's can be merged as one. So your final string will look something like (a...a)b(a...a)b(a...)...

You can now consider this problem as sum of all possible product of subsets of a given set, where each element in the set is the number of a's delimited by b.

For example: In the string "aaabaabaaab", set formed will be {3,2,3,0} (0 can be ignored). Now if you have a set {a1,a2,...,aN}, then sum of all possible products of this set is equal to (1+a1)(1+a2)...*(1+aN)-1.

Proof:

Write the required answer as follows:

S = Sum of products of subset with (size=1)+(size=2)+...(size=N)

\(S = (a_1+a_2+...a_N)+(a_1*a_2+a_1*a_3......+a_{N-1}*a_N)+...+(a_1*a_2.....a_N)\)

After factorization,

S = (1+a1)(1+a2)...(1+aN)-1

大致意思就是处理所给的字符串，，，就变成了一堆a一个b一堆a一个b....这样的，，，

也就是一堆a的集合

题目所要的就是调两个个a的集合里调一个a作为子串的首尾，，，这样的就是所要的串，，，

于是总共的个数就是\(a_1*a_2*a_3.....a_{n-1}*a_{n}\),,,对了每个集合还要加一，，表示这个集合选一个或者都不选，，，最后的答案再减一就行了（全不选的情况不符合题意），，，

他上面那段话的思路是在计算所有的子川的情况时，，长度为1+长度为2+。。。长度为m。。。

这样的话 \(sum = (a_1 + a_2+....+a_{n-1} + a_n) + (a_1 * a_2 + a_1 * a_3 + ...+ a_1 * a_m + a_2 * a_3 + ......+ a_{n-1} * a_n) + (a_1 * a_2 * a_3 + ....) + ...\)

然后这个求和可以转化成\(sum = (1 + a_1) * (a_2 + 1) * (a_3 + 1) + (a_4 + 1) * .... * (a_n + 1) - 1\)

因为：

\(a + b + ab = (a + 1) * (b + 1) - 1\)

\(a + b + c + ab + ac + bc + abc = (a + 1) * (b + 1) * (c + 1) - 1\)

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;

ll a[maxn];
int main()
{
//    freopen("233.txt" , "r" , stdin);
//    freopen("233.out" , "w" , stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int n = 0;
    string s;cin >> s;
    int a[maxn];
    int tot = 0;
    int len = s.length();
    for(int i = 0; i < len; ++i)
    {
        if(s[i] == 'a')
        {
            int cnt = 0;
            for(int j = i; j < len; ++j)
            {
                if(s[j] == 'a')
                    ++cnt;
                if(s[j] == 'b' || j == len - 1)
                {
                    a[tot++] = cnt + 1;
                    i = j;
                    break;
                }

            }
        }
    }
    ll ans = 1;
    for(int i = 0; i < tot; ++i)
        ans = (ans * a[i]) % mod;
    --ans;
    cout << ans << endl;
    return 0;
}

太水了，，，，QAQ