Educational Codeforces Round 105 (Rated for Div. 2) AB题解

A. ABC String

思路：相同字符要有相同的半括号（要么都是左括号要么都是右括号），总共8种情况。若把左括号看做1，右括号看成-1，那么这个序列满足任意前缀和\(sum[i]>=0且sum[n]==0\)，对每种情况进行检验即可。

view code

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        string s;
        cin>>s;
        bool ok = false;
        for(int A=1, cnt1 = 1; cnt1<=2; cnt1++, A *= -1)
        for(int B=1, cnt2 = 1; cnt2<=2; cnt2++, B *= -1)
        for(int C=1, cnt3 = 1; cnt3<=2; cnt3++, C *= -1)
        {
            int sum[55];
            mem(sum,0);
            map<char, int> Map;
            if(A==1) Map['A'] = 1;
            else Map['A'] = -1;
            if(B==1) Map['B'] = 1;
            else Map['B'] = -1;
            if(C==1) Map['C'] = 1;
            else Map['C'] = -1;
            string t = s;
            for(int i=0; i<t.size(); i++) sum[i+1] = sum[i] + Map[t[i]];
            int flag = 1;
            for(int i=1; i<=t.size(); i++)
            {
                if(sum[i]<0||sum[t.size()] != 0)
                {
                    flag = 0;
                    break;
                }
            }
            if(flag)
            {
                ok = true;
                break;
            }
        }
        puts(ok?"YES":"NO");
    }
    return 0;
}

B. Berland Crossword

思路：其实各边能产生关系的也就四个顶点。所以枚举一下四个顶点有无填黑的情况（我这里采用四位二进制表示），看看剩下的n-2个位置能否满足对应数量的黑块即可。

view code

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[10];

int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        ll n = read();
        rep(i,1,4) a[i] = read();
        int flag = 0;
        for(ll i=0; i<16; i++)
        {
            int TopLeft = (i&1);
            int TopRight = (i>>1)&1;
            int BottomLeft = (i>>2)&1;
            int BottomRight = (i>>3)&1;
            int U = a[1];
            int R = a[2];
            int D = a[3];
            int L = a[4];
            U -= TopLeft;
            U -= TopRight;
            R -= TopRight;
            R -= BottomRight;
            D -= BottomRight;
            D -= BottomLeft;
            L -= BottomLeft;
            L -= TopLeft;
            if(U>=0&&R>=0&&D>=0&&L>=0&&U<=n-2&&R<=n-2&&D<=n-2&&L<=n-2)
            {
                flag = 1;
                break;
            }
        }
        puts(flag?"YES":"NO");
    }
    return 0;
}