Multi-University 2015 #6 E(hdu 5357 Easy Sequence)
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题目链接
E - Easy Sequence
Time Limit:1000MS Memory Limit:131072KB
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≤|s|≤106).
Output
For each test case, output an integer m=∑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,
then
题目大意
给定一个括号序列,令ans[i]表示包含第i个字符的合法的字符串个数。
求
题解
对于一个合法的括号串,都可以化成这样的形式
定义
当然,a仅对str[i]==’(‘时有效,b仅对str[i]==’)’。
于是
这样就代表了两种情况:
里面的两个匹配的括号被外面的包含在内,
后面是一个独立的匹配括号。
对于第一种情况,a[i]当然是1,而由于match[i]+1是’)’所以
而对于第二种情况,
b的推导也与此类似。
从这个推导式可以看出来,a数组需要倒着求,b数组需要正着求。
再考虑如何求出
其中
对于这个表达式,可以这样理解:以a开头的合法串有
这题非常坑的一点就是这个取模是在
对于match,up这些数组,可以用一个栈来算出,值得注意的是,计算up的时候,要先判这个左括号是不是有匹配的右括号,有才会计算up再塞入栈中。
复杂度为
#include<cstdio>#include<cstring>const int M=1000005;const int mod=1e9+7;typedef long long ll;int match[M],stk[M],up[M],a[M],b[M],ans[M];char str[M];void solve(){ scanf("%s",str+1); int len=strlen(str+1),top=0; for(int i=0;i<=len+1;i++) up[i]=a[i]=b[i]=ans[i]=match[i]=0; for(int i=1;i<=len;i++){ if(str[i]=='('){ stk[++top]=i; }else if(top){ match[match[stk[top]]=i]=stk[top]; top--; } } for(int i=len;i>=1;i--) if(str[i]=='('&&match[i]) a[i]=a[match[i]+1]+1; for(int i=1;i<=len;i++) if(str[i]==')'&&match[i]) b[i]=b[match[i]-1]+1; top=0; for(int i=1;i<=len;i++){ if(str[i]=='('&&match[i]){ while(top&&match[stk[top]]<match[i]) top--; up[i]=stk[top]; stk[++top]=i; } } for(int i=1;i<=len;i++) if(str[i]=='(') ans[i]=ans[match[i]]=(ans[up[i]]+a[i]*1LL*b[match[i]])%mod; ll sum=0; for(int i=1;i<=len;i++) sum=sum+1LL*i*ans[i]%mod; printf("%lld\n",sum);}int main(){ int cas; scanf("%d",&cas); while(cas--) solve(); return 0;}
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