10.11NOIP模拟题(2)

 

/*
string水过
*/
#include<bits/stdc++.h>

#define N 1001

using namespace std;
int n,x,y,m,pre;
string str;
string s[N];

inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int main()
{
    freopen("y.in","r",stdin);
    freopen("y.out","w",stdout);
    n=read();
    for(register int i=1;i<=n;i++) cin>>s[i];
    str="I_love_";
    m=read();
    for(register int i=1;i<=m;i++)
    {
        x=read();y=read();
        s[x]=str;s[x]+=s[y];pre=x;
    }
    cout<<s[1];
    return 0;
}

 

 

/*
ai<=ai+1时
枚举每一个区间长度二分右端点
*/
#include<bits/stdc++.h>

#define N 500001
#define ll long long

using namespace std;
ll n,ans,K,m;
ll a[N],sum[N];

inline ll read()
{
    int x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

void work1()
{
    while(m--)
    {
        K=read();ans=0;
        for(int j=1;j<=n;j++)
        {
            for(int k=j;k<=n;k++)
            {
                ll res1=sum[k]-sum[j-1];
                ll res2=k-j+1;
                if(res1<=K*res2) ans++;
            }
        }
        cout<<ans<<endl;
    }    
}

void work2()
{
    ll l,r,mid,tmp,lim;
    while(m--)
    {
        K=read();ans=0;lim=1;tmp=0;
        for(int L=1;L<=n;L++)
        {
            l=1;r=n+1;lim=L*K;
            while(l<=r)
            {
                mid=(l+r)>>1;
                if(sum[mid]-sum[mid-L]<=lim && sum[mid]) 
                  tmp=mid,l=mid+1;
                else r=mid-1;
            }
            if(tmp-L+1>0)
              ans+=tmp-L+1;
        }
        cout<<ans<<endl;
    }
}

int main()
{
    freopen("h.in","r",stdin);
    freopen("h.out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++) a[i]=read();
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
    m=read();
    if(n<=100) work1();
    else work2();
    return 0;
}

/*
将数列减k转化为求区间和<=0的区间个数
前缀和，求多少l,r满足s[r]-s[l-1]<=0
即s[r]<=s[L]（L=l-1）
树状数组求逆序对
*/
#include<bits\stdc++.h>

#define N 50001
#define ll long long

using namespace std;
ll n,m,k;
ll ori[N],a[N],b[N],tmp[N];

inline int discrete()
{
    for(int i=1;i<=n;i++) tmp[i]=a[i];
    sort(tmp+1,tmp+n+1);
    ll *end=unique(tmp+1,tmp+n+1);
    for(int i=1;i<=n;i++) a[i]=lower_bound(tmp+1,end,a[i])-tmp;
    return end-(tmp+1);
}

struct BinaryIndexedTree
{
    int n; ll a[N];
    
    static int lowbit(int x)
    {
        return x & -x;
    }
    
    void clear()
    {
        for(int i=1;i<=n;i++) a[i]=0;
    }
    
    ll query(int pos)
    {
        ll ans=0;
        for(int i=pos;i>0;i-=lowbit(i)) ans+=a[i];
        return ans;
    }
    
    void update(int pos,int del)
    {
        for(int i=pos;i<=n;i+=lowbit(i)) a[i]+=del;
    }
    
}bit;

inline ll read()
{
    ll x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

inline ll solve(int k)
{
    ll ans=0;
    for(int i=1;i<=n;i++) 
    {
        a[i]=ori[i]-k+a[i-1];
        if(a[i]<=0) ans++;
    }
    bit.n=discrete();
    bit.clear();
    
    for(int i=n;i>=1;i--)
    {
        ans+=bit.query(a[i]);
        bit.update(a[i],1);
    }
    return ans;
}

int main()
{
    freopen("h.in","r",stdin);
    freopen("h.out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++) ori[i]=read();
    m=read();
    while(m--)
    {
        k=read();
        printf("%lld\n",solve(k));
    }
}

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

#define N 3001

using namespace std;
int n,m,T,flag;
char ch;
bool e1[N][N],e2[N][N];

void floyed()
{
    flag=0;
    for(int k=1;k<=n;k++) for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
          if(i!=j && j!=k)
          {
              if(e1[i][k] && e1[k][j] && !e1[i][j]) flag=1;
              if(flag) break;
              if(e2[i][k] && e2[k][j] && !e2[i][j]) flag=1;
              if(flag) break;
          }
        if(flag) break;
    }
}

int main()
{
    freopen("p.in","r",stdin);
    freopen("p.out","w",stdout);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(e1,0,sizeof e1);
        memset(e2,0,sizeof e2);
        for(int i=1;i<=n;i++) for(int j=1;j<=n;j++)
        {
            cin>>ch;
            if(ch=='H') e1[i][j]=1;
            if(ch=='W') e2[i][j]=1;    
        }
        floyed();
        if(flag) printf("N0\n");
        else printf("YE5\n");
    }
    return 0;
}

 

/*
考虑u->v和v->w都被一个人占领
那么答案为N0的情况为
1.w->u 另一人占领
2.u->w 另一人占领
3.w->u 同一人占领
1,3有环 
2 将一个人反向建图有环
所以两次拓扑排序判断是否有换即可
*/
#include<bits\stdc++.h>

#define N 3001

using namespace std;
int n,m,k,ans,flag;
char s[N][N];
int e[N][N],in[N];

void T_sort()
{
    memset(in,0,sizeof in);
    for(int i=1;i<=n;i++) for(int j=1;j<=n;j++)
    if(e[i][j]) in[j]++;
    for(int i=1;i<=n;i++)
    {
        int u=0;
        for(int j=1;j<=n;j++)
        {
            if(in[j]==0)
            {
                u=j;in[j]=-1;break;
            }
        }
        if(!u){flag=1;return;}
        for(int j=1;j<=n;j++) if(e[u][j]) in[j]--;
    }
}

int main()
{
    freopen("p.in","r",stdin);
    freopen("p.out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        flag=0;
        for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
        for(int k=0;k<=1;k++) 
        {
            memset(e,0,sizeof e);
            for(int i=1;i<=n;i++) for(int j=1;j<=n;j++)
            {
                if(s[i][j]=='H') e[i][j]=1;
                if(s[i][j]=='W')
                {
                    if(!k) e[i][j]=1;
                    else e[j][i]=1;
                }
            }
            T_sort();
        }
        if(!flag) printf("YE5\n");
        else printf("N0\n");
    }
}