牛客 201 J Princess Principal (括号, 栈模拟)

大意: 给定序列$a$, $a_i$为偶数代表第$\frac{a_i}{2}$种左括号, 否则为第$\frac{a_i-1}{2}$种右括号. 询问区间是否是合法括号序列.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#include <unordered_map>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
template <class T> void rd(T &x){x=0;bool f=0;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=1;c=getchar();}while('0'<=c&&c<='9'){x=x*10+c-'0';c=getchar();}if(f)x=-x;}
//head


const int N = 1e6+10;
int n,m,q,a[N],s[N],L[N];

int main() {
	scanf("%d%d%d", &n, &m,&q);
	REP(i,1,n) scanf("%d",a+i);
	int top = 0;
	REP(i,1,n) {
		if (!top) s[++top] = i;
		else {
			if (a[i]/2==a[s[top]]/2&&a[i]==a[s[top]]+1) --top;
			else s[++top] = i;
		}
		L[i] = top;
	}
	REP(i,1,q) {
		int x, y;
		scanf("%d%d", &x, &y);
		puts((x-y&1)&&L[x-1]==L[y]?"YES":"NO");
	}
}

标签：201,a%,括号,ll,牛客,return,include,Princess,define
来源： https://www.cnblogs.com/uid001/p/11025771.html