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 = 1000000007;
const int maxn = 2000010;
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 + 1);
        top = 0;
        n = strlen(str + 1);
        for(int i = 1; i <= n; ++i) {
            match[i] = pre[i] = 0;
            if(str[i] == '(') stk[++top] = i;
            else if(top) {
                match[stk[top]] = i;
                match[i] = stk[top];
                if(top > 1) pre[match[i]] = stk[top-1];
                stk[top--] = 0;
            }
        }
        ans[0] = a[0] = b[n+1] = 0;
        for(int i = 1; i <= n; i++)
            a[i] = (str[i] == ')' && match[i])?(a[match[i] - 1] + 1):0;
        for(int i = n; i >= 1; i--)
            b[i] = (str[i] == '(' && match[i])?(b[i] = b[match[i] + 1] + 1):0;
        for(int i = 1; i <= n; i++)
            ans[i] = (str[i] == '(')?(ans[pre[i]] + ((LL)b[i]*a[match[i]] % mod) % mod):ans[match[i]];
        LL ret = 0;
        for(int i = 1; i <= n; ++i)
            ret += ans[i]*i%mod;
        printf("%I64d
",ret);
    }
    return 0;
}