Time Limit: 10 Sec Memory Limit: 128 MB
Description
Recently, the cows have been competing with strings of balanced
parentheses and comparing them with each other to see who has the
best one.
Such strings are scored as follows (all strings are balanced): the
string "()" has score 1; if "A" has score s(A) then "(A)" has score
2*s(A); and if "A" and "B" have scores s(A) and s(B), respectively,
then "AB" has score s(A)+s(B). For example, s("(())()") =
s("(())")+s("()") = 2*s("()")+1 = 2*1+1 = 3.
Bessie wants to beat all of her fellow cows, so she needs to calculate
the score of some strings. Given a string of balanced parentheses
of length N (2 <= N <= 100,000), help Bessie compute its score.
计算“平衡字符串”的分数，“平衡字符串”是指由相同数量的‘（’和‘）’组成，
且以‘（’开头，以‘）’结尾的字符串。
计算规则：
字符串“（）”的得分是1.
如果，平衡字符串“A”的得分是是S(A)，那么字符串“（A）”得分是2*S(A) ；
如果，“A”，“B” 得分分别是S(A)和S(B)，那么平衡字符串“AB”得分为S(A)+S(B)
例如：s("(())()") =s("(())")+s("()") = 2*s("()")+1 = 2*1+1 = 3.
Input
* Line 1: A single integer: N
* Lines 2..N + 1: Line i+1 will contain 1 integer: 0 if the ith
character of the string is '(', and 1 if the ith character of the string is ')'
第1行：N，平衡字符串长度
第2至N+1行：Linei+1 整数0或1,0代表字符‘（’，1代表‘）’
Output
* Line 1: The score of the string. Since this number can get quite large, output the score modulo 12345678910.
计算字符串得分，结果对12345678910取模
Sample Input
6
0
0
1
1
0
1
INPUT DETAILS:
This corresponds to the string "(())()".
Sample Output
3
HINT
Source
Silver
解：
期望：100 实际：20
暴力实现有问题。读题不认真，没有取模。太菜了。
我的想法：从小括号开始扫，每次将对应包含的一个括号集团消除掉
不知道哪里写挂了。
为什么我总是这么弱呢？
正解：分治。
预处理出每个左括号所对应的右括号的所在位置。
然后对于整个问题，分治，用搜索得到部分解，然后更新答案。
1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 #define ll long long 5 using namespace std; 6 const ll mod=12345678910ll; 7 int n,s[200000],top,pos[200000]; 8 ll dfs(int l,int r) 9 { 10 ll nw=0,p=pos[l]; 11 if(l+1==p) nw=(nw+1)%mod; 12 else nw=(nw+2*dfs(l+1,p-1)%mod)%mod; 13 if(p+1<=r) nw=(nw+dfs(p+1,r)%mod)%mod; 14 return nw; 15 } 16 int main() 17 { 18 scanf("%d",&n); 19 for(int i=1,x;i<=n;++i) 20 { 21 scanf("%d",&x); 22 if(x==0) s[++top]=i; 23 else if(top) pos[s[top--]]=i; 24 } 25 printf("%lld",dfs(1,n)); 26 return 0; 27 }