• 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 '
    '
    #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 = 4e6+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] = s[top];
    	}
    	REP(i,1,q) {
    		int x, y;
    		scanf("%d%d", &x, &y);
    		puts(L[x-1]==L[y]?"Yes":"No");
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/11025771.html
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