2015 Multi-University Training Contest 6 hdu 5357 Easy Sequence

Easy Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 766    Accepted Submission(s): 208

Problem Description
soda has a string containing only two characters -- '(' and ')'. For every character in the string, soda wants to know the number of valid substrings which contain that character.

Note:
An empty string is valid. If S is valid, (S) is valid. If U,V are valid, UV is valid.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

A string s consisting of '(' or ')' $(1 \leq |s| \leq 10^6)$.

Output
For each test case, output an integer $m=\sum_{i=1}^{|s|}(i⋅ansi mod 1000000007)$, where ansi is the number of valid substrings which contain i-th character.

Sample Input
2
()()
((()))

Sample Output
20
42

Hint

For the second case, $ans = \{1, 2, 3, 3, 2, 1\}$, then $m=1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1 = 42$

Author
zimpha@zju

Source

2015 Multi-University Training Contest 6 1005

 

解题：栈

 

貌似还是有点不明白，代码先放着

 

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const LL mod = ;
 const int maxn = ;
 char str[maxn];
 int stk[maxn],match[maxn],pre[maxn],a[maxn],b[maxn],top;
 LL ans[maxn];
 int main() {
     int kase,n;
     scanf("%d",&kase);
     while(kase--) {
         scanf("%s",str + );
         top = ;
         n = strlen(str + );
         for(int i = ; i <= n; ++i) {
             match[i] = pre[i] = ;
             if(str[i] == '(') stk[++top] = i;
             else if(top) {
                 match[stk[top]] = i;
                 match[i] = stk[top];
                 if(top > ) pre[match[i]] = stk[top-];
                 stk[top--] = ;
             }
         }
         ans[] = a[] = b[n+] = ;
         for(int i = ; i <= n; i++)
             a[i] = (str[i] == ')' && match[i])?(a[match[i] - ] + ):;
         for(int i = n; i >= ; i--)
             b[i] = (str[i] == '(' && match[i])?(b[i] = b[match[i] + ] + ):;
         for(int i = ; i <= n; i++)
             ans[i] = (str[i] == '(')?(ans[pre[i]] + ((LL)b[i]*a[match[i]] % mod) % mod):ans[match[i]];
         LL ret = ;
         for(int i = ; i <= n; ++i)
             ret += ans[i]*i%mod;
         printf("%I64d\n",ret);
     }
     return ;
 }