山东省赛补题

Posted 2022-03-19 钟钟终

Games

https://ac.nowcoder.com/acm/contest/30256/G
前i堆石子取出j堆异或和为sum，使得sum^sum=0,使得后手必胜。
每堆石子的数量最大为1000，极端情况会出现二进制10个1，最大为1023，k最大取1023.

#include <bits/stdc++.h>
#define long long ll
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=3e9+7;
int f[1001][11][1024],a[1001];
signed main()

    int t;scanf("%d",&t);
    while(t--)
    

        memset(f,0,sizeof(f));
        int n,d;scanf("%d%d",&n,&d);
        int sum=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),sum^=a[i];
        for(int i=0;i<=n;i++)
            f[i][0][0]=1;
        for(int i=1;i<=n;i++)
        
            for(int j=1;j<=d;j++)
                for(int k=0;k<1024;k++)
                f[i][j][k]=f[i-1][j-1][k^a[i]]+f[i-1][j][k];
                f[i][j][k]%=mod;
                
        
        int ans=0;
        for(int j=0;j<=d;j++)
            ans+=f[n][j][sum],ans%=mod;
        cout<<ans%mod<<endl;
    
    return 0;

A - Calandar

由于一年的天数和一个月的天数，取模于5都等于 0，所以只要考虑天数即可。若天数小于前面，则加上30取模5。根据周几加上天数，将5和0等价。

#include <bits/stdc++.h>
#define long long ll
using namespace std;
const int inf=0x3f3f3f3f;
string s[5]="Friday","Monday","Tuesday","Wednesday","Thursday";
signed main()

    int t;scanf("%d",&t);
    while(t--)
    
        int y1,m1,d1,y2,m2,d2;string ss;
        cin>>y1>>m1>>d1>>ss>>y1>>m2>>d2;
        int tmp=(d2+30-d1)%5,ans;  
        if(ss=="Monday")
            ans=(tmp+1)%5;
        else if(ss=="Tuesday")
            ans=(tmp+2)%5;
        else if(ss=="Wednesday")
            ans=(tmp+3)%5;
        else if(ss=="Thursday")
            ans=(tmp+4)%5;
        else
            ans=(tmp+5)%5;
        cout<<s[ans]<<endl;
    
    return 0;

H Dominoes

空格移动的方向要和将要移动的骨牌另一端的格子相一致

#include <bits/stdc++.h>
#define long long ll
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn=1e5+5;
int dx[5]=1,0,-1,0;
int dy[5]=0,1,0,-1;
map<pair<int,int>,pair<int,int> >mp;
struct node

    int x,y;
cur;
bool vis[12][maxn];
signed main()

    int n,m,k;
    while(cin>>n>>m>>k)
    mp.clear();memset(vis,0,sizeof(vis));
    for(int i=1;i<=k;i++)
    
        int a,b,c,d;cin>>a>>b>>c>>d;
        mp[a,b]=c,d;mp[c,d]=a,b;
        vis[a][b]=1;vis[c][d]=1;
    
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        if(!vis[i][j])
        
            cur.x=i;cur.y=j;break;
        
    memset(vis,0,sizeof(vis));
    int ans=0;
    //cout<<cur.x<<" "<<cur.y<<endl;
    queue<node>q;q.push(cur);
    while(!q.empty())
    
        cur=q.front();q.pop();
        int xx=cur.x,yy=cur.y;
        if(vis[xx][yy])
            continue;
        vis[xx][yy]=1;
        for(int i=0;i<4;i++)
        
            int x1=xx+dx[i],y1=yy+dy[i];
            int x2=mp[x1,y1].first,y2=mp[x1,y1].second;
            if(x1>=1&&x1<=n&&y1>=1&&y1<=m&&abs(x2-x1)==abs(dx[i])&&abs(y2-y1)==abs(dy[i])&&!vis[x2][y2])
            
                q.push(nodex2,y2);
                ans++;

            
        
    
    cout<<ans<<endl;
    
    return 0;

F - Stones in the Bucket

思维题，虽然比较简单，但是思路挺清晰的

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn=1e5+5;
int a[maxn];
signed main()

    int t;cin>>t;
    while(t--)
    
        int n;cin>>n;
        int sum=0;
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]),sum+=a[i];
        int tmp=sum/n,s1=0,s2=0;
        for(int i=1;i<=n;i++)
        
            if(a[i]<tmp)
                s1+=tmp-a[i];
            else if(a[i]>tmp)
                s2+=a[i]-tmp;
        
        cout<<s1+(s2-s1)<<endl;
    
    return 0;

H - Tokens on the Segments

对于我的难点在于优先队列和结构体结合的排序上，没用过几次，掌握的不好。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn=1e5+5;
struct node

    int x,y;
    bool operator <(const node &a)const
    
        if(a.x==x)
            return a.y<y;
        return a.x<x;
    
e[maxn];
signed main()

    int t;cin>>t;
    while(t--)
    
        priority_queue<node>q;
        int n;cin>>n;
        for(int i=1;i<=n;i++)
        
            scanf("%lld%lld",&e[i].x,&e[i].y);
            q.push(e[i]);
        
        int mx=0,ans=0;
        while(!q.empty())
        
            node cur=q.top();q.pop();
            if(cur.x>mx)
            
                ans++;
                mx=cur.x;
            
            else if(cur.x<=mx&&cur.x+1<=cur.y)
            
                cur.x++;
                q.push(cur);
            
        
        cout<<ans<<endl;
    
    return 0;