Codeforces Round #662 (Div. 2) B. Applejack and Storages(思维)

题目链接

思路:

记录个数超过2,4,6,8,的木板的个数,如果他们之间的逻辑满sum4>0&&sum2>2||sum6>0&&sum2>1||sum8>0||sum4>1就输出YES,否则就输出NO。

代码:

#include<bits/stdc++.h>
//#define int long long
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int N=1e5+3;
const int M=2e4+5;
const double eps=1e-8;
const int mod=998244353;
const int inf=0x7fffffff;
const double pi=3.1415926;
using namespace std;
int flag[N];
signed main()
{
    IOS;
    memset(flag,0,sizeof flag);
    int n,m,sum2=0,sum4=0,sum6=0,sum8=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        int a;
        cin>>a;
        flag[a]++;
        if(flag[a]==2)
        {
            sum2++;
        }
        else if(flag[a]==4)
        {
            sum4++;
        }
        else if(flag[a]==6)
        {
            sum6++;
        }
        else if(flag[a]==8)
        {
            sum8++;
        }
    }
    cin>>m;
    for(int i=0;i<m;i++)
    {
        char c;
        int a;
        cin>>c>>a;
        if(c=='+')
        {
            flag[a]++;
            if(flag[a]==2)
            {
                sum2++;
            }
            else if(flag[a]==4)
            {
                sum4++;
            }
            else if(flag[a]==6)
            {
                sum6++;
            }
            else if(flag[a]==8)
            {
                sum8++;
            }
        }
        else
        {
            flag[a]--;
            if(flag[a]==1)
            {
                sum2--;
            }
            else if(flag[a]==3)
            {
                sum4--;
            }
            else if(flag[a]==5)
            {
                sum6--;
            }
            else if(flag[a]==7)
            {
                sum8--;
            }
        }
        if(sum4>0&&sum2>2||sum6>0&&sum2>1||sum8>0||sum4>1)
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/ACkingdom/article/details/107873685