[CSP-S Simulation Test]: Simple brackets sequence (number of combinations)

Topic Portal (internal title 82)

Input Format

　　A string line $ ss $, $ SS $ ensure contains only $ '(' $ and $ ')' $.

Output Format

　　A line integer representing the number of sub-sequences satisfy requirements result $ 10 $ ^ 7 + 9 in.

Sample

Sample input 1:

)(()()

Sample output 1:

Sample input 2:

()()()

Sample Output 2:

Sample input 3:

)))

Sample output 3:

Data range and tips

Sample explained:

　　The first set of examples, there are several sub-sequences satisfy condition (subscript string count from $ 1 $):
　　delete characters $ $ 1,5 positions to obtain $ (()) $
　　remove $ 1,2,3,4 $ character position to obtain $ () $
　　delete $ $ 1,2,4,5 position of the character, get $ () $
　　delete $ $ 1,2,5,6 character position to obtain $ () $
　　delete $ 1,3 , $ 4,5 character position to obtain $ () $
　　delete characters $ $ 1,3,5,6 position to obtain $ () $

data range:

　　Provided $ n-$ is $ SS $ length
　　data for $ 20 \% $ a, $ n leqslant 20 $ \
　　to $ 50 \% $ data, $ n \ leqslant 2,000 $
　　for $ 100 \% $ data, $ n \ leqslant 200,000 $

answer

From left to right around the demarcation point you can enumerate the brackets, but will obviously re-count, consider the inclusion and exclusion?

In fact, there is a simpler approach to directly request the left (right) side of the brackets must be selected just fine.

Then there is the $ n ^ 2 $ approach.

But if you have done this question: permutations and combinations , it is much simpler.

Time complexity: $ \ Theta (n) $.

Expectations score: $ 100 $ points.

Actual score: $ 100 $ points.

Code time

#include<bits/stdc++.h>
using namespace std;
const int mod=1000000007;
int n;
char ch[200001];
int l[200001],r[200001];
long long fac[200001],inv[200001];
long long ans;
long long qpow(long long x,long long y)
{
	long long res=1;
	while(y)
	{
		if(y&1)res=res*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return res;
}
void pre_work()
{
	fac[0]=1;
	for(int i=1;i<=200000;i++)
		fac[i]=fac[i-1]*i%mod;
	inv[200000]=qpow(fac[200000],mod-2);
	for(int i=200000;i;i--)
		inv[i-1]=inv[i]*i%mod;
}
long long C(long long x,long long y)
{
	if(x<y)return 0;
	return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int main()
{
	pre_work();
	scanf("%s",ch+1);
	n=strlen(ch+1);
	for(int i=1;i<=n;i++)l[i]=l[i-1]+(ch[i]=='(');
	for(int i=n;i;i--)r[i]=r[i+1]+(ch[i]==')');
	for(int i=1;i<=n;i++)
		if(ch[i]=='(')
			ans=(ans+C(l[i-1]+r[i+1],r[i+1]-1))%mod;
	printf("%lld",ans);
	return 0;
}

rp++