2018牛客网多校训练6

https://www.nowcoder.com/acm/contest/144#question

凉了啊,四题最后一名

D:水题

solved by lyy

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t,n,m,k;
int v[100005];
 
int main()
{
    scanf("%d",&t);
    int step=0;
    while (t--)
    {
        step++;
        memset(v,0,sizeof(v));
        scanf("%d%d%d",&n,&m,&k);
        while (k--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            v[b]=max(v[b],c);
        }
        ll ans=0;
        for (int i=1;i<=m;i++)
        {
            ans+=v[i];
        }
        printf("Case #%d: %lld\n",step,ans);
    }
    return 0;
}

A:

solved by wyq

#include <bits/stdc++.h>
using namespace std;
#define ll long long
queue<int> q;
int a[20005][20];
int main()
{
    int T,i,x,y,n,len,max1,max2,j,count=0;
    scanf("%d",&T);
    memset(a,-1,sizeof(a));
    while(T--)
    {
        count++;
        scanf("%d",&n);
        len=1<<n;
        for(i=1;i<=len;i++)
        {
            for(j=1;j<=n;j++)
            scanf("%d",&a[i][j]);
        }
        for(i=1;i<=len;i++)
        {
            sort(a[i]+1,a[i]+n+1);
            q.push(i);
        }
        while(true)
        {
            //printf("%d\n",q.size());
            if(q.size()==1)break;
            x=q.front();q.pop();
            y=q.front();q.pop();
            //printf("%d %d\n",x,y);
            max1=0;
            for(i=n;i>=1;i--)
            {
                if(a[x][i]==-1)continue;
                if(a[x][i]>max1)max1=a[x][i];
            }
            max2=0;
            for(i=n;i>=1;i--)
            {
                if(a[y][i]==-1)continue;
                if(a[y][i]>max2)max2=a[y][i];
            }
            if(max1>max2)
            {
                //printf("%dwin\n",x);
                q.push(x);
                for(i=1;i<=n;i++)
                {
                    if(a[x][i]==-1)continue;
                    if(a[x][i]>max2)
                    {
                        a[x][i]=-1;
                        break;
                    }
                }
            }
            else
            {
                //printf("%dwin\n",y);
                q.push(y);
                for(i=1;i<=n;i++)
                {
                    if(a[y][i]==-1)continue;
                    if(a[y][i]>max1)
                    {
                        a[y][i]=-1;
                        break;
                    }
                }
            }
        }
        printf("Case #%d: %d\n",count,q.front());
        q.pop();
    }          
    return 0;
}

J:瞎搞过的,发现题解说的很有道理啊,结论知道,但是当时没想到,方法也跟这个差不多吧

solved by lyy

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxx = 1e7+2;
vector<unsigned int> v;
 
int main()
{
    int n,T;
    scanf("%d",&T);
    int cnt=0;
    while(T--)
    {
        v.clear();
        int a,b,c;
        scanf("%d%u%u%u",&n,&a,&b,&c);
        unsigned int x = a,y = b,z = c;
        for(int i = 0; i < n; i++)
        {
            unsigned int t;
            x ^= x << 16;
            x ^= x >> 5;
            x ^= x << 1;
            t = x;
            x = y;
            y = z;
            z = t ^ x ^ y;
            v.push_back(z);
        }
        //sort(v.begin(),v.end());
        unsigned long long nowmax=0,now;
        if (v.size()<=1000000)
        {
            for (int i=0;i<min(2,n);i++)
            {
                for (int j=i+1;j<n;j++)
                {
                    if (v[i]<v[j]) swap(v[i],v[j]);
                }
            }
             for(int i=0;i<n;i++)
            {
                if(v[i]<nowmax/v[i]) continue;
                for(int j=0;j<n;j++)
                {
                    if(v[j]<nowmax/v[i] || i==j) continue;
                    now=(unsigned long long)v[i]*v[j]/(unsigned long long)__gcd(v[i],v[j]);
                    nowmax=max(now,nowmax);
                }  
            }
        }
        else
        {
            while (v.size()>1000000)
            {
                vector<unsigned int> vv;
                unsigned int r=max(v[100],v[200]);
                for (auto x:v)
                {
                    if (x>r) vv.push_back(x);
                }
                v=vv;
                //cout<<v.size()<<endl;
            }
            for (int i=0;i<min(2,(int)v.size());i++)
            {
                for (int j=i+1;j<v.size();j++)
                {
                    if (v[i]<v[j]) swap(v[i],v[j]);
                }
            }
             for(int i=0;i<v.size();i++)
            {
                if(v[i]<nowmax/v[i]) continue;
                for(int j=0;j<v.size();j++)
                {
                    if(v[j]<nowmax/v[i] || i==j) continue;
                    now=(unsigned long long)v[i]*v[j]/(unsigned long long)__gcd(v[i],v[j]);
                    nowmax=max(now,nowmax);
                }  
            }
        }
        printf("Case #%d: %llu\n",++cnt,(unsigned long long)nowmax);
    }
     
    return 0;
}

C:就是推推公式,没有预先处理逆元会TLE(我就是这样T的)

solved by lyy

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=998244353;
ll n,m;
int t;
 
long long inv(long long a,long long m)
{
    if(a == 1)return 1;
    return inv(m%a,m)*(m-m/a)%m;
}
 
ll f[1000005];
 
int main()
{
    for (int i=1;i<=1000000;i++)
    {
        f[i]=inv(i,mod);
    }
    cin>>t;
    int step=0;
    while (t--)
    {
        step++;
        cin>>n>>m;
//      if (m==1)
//      {
//          printf("Case #%d: %lld\n",step,1);
//          continue;
//      }
        int p=(int)min(n,m);
        ll x=1;
        ll y=m%mod;
        ll ans=0;
        for (int i=1;i<=p;i++)
        {
            //cout<<x<<' '<<y<<endl;
            ans=(ans+x*y)%mod;
            if (i!=p)
            {
                y=y*((m-i)%mod)%mod;
                x=x*((n-i)%mod)%mod*f[i]%mod;
            }
        }
        printf("Case #%d: %lld\n",step,ans);
    }
    return 0;
}

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转载自blog.csdn.net/njupt_lyy/article/details/81437730