Bestcoder round#84 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/xaphoenix/article/details/52087831

本文解析了五道涉及模拟、最长递增子序列、二维数组处理、质数判断及区间查询等算法问题的编程题，提供了完整的代码实现，并对关键步骤进行了说明。

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1001 Aaronson

简单模拟

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;
int T;
int x,y,z,zz;
int flag;
int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d %d",&x,&z);
        zz=1;
        flag=0;
        for (int i=1;i<=z;i++)
        {
            zz*=2;
            if (zz>=x)
            {
                flag=1;
                break;
            }
        }
        if (flag) y=0;
        else y=x/zz,x-=x/zz*zz;
        while (x) y+=x&1,x>>=1;
        printf("%d\n",y);
    }
    return 0;
}

1002 Bellovin

很容易发现，字典序最小的答案就是 $f$ 数组的值，所以此题就是裸的 $nlogn$ 求LIS

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;

int T,n;
int a[410000],f[410000];
int c[410000],tail;
int check(int x)
{
    if (x>c[tail]) c[++tail]=x;
    int l=1,r=tail,ans=0;
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (c[mid]<x) ans=mid,l=mid+1;
        else r=mid-1;
    }
    return ans;
}
int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        memset(c,0,sizeof(c));
        tail=0;
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if (i==1) c[++tail]=a[i],f[1]=1;
            else
            {
                int pos=check(a[i]);
                f[i]=pos+1;
                if (c[pos+1]>a[i]) c[pos+1]=a[i];
            }
        }

        for (int i=1;i<=n;i++)
        {
            printf("%d",f[i]);
            if (i!=n) printf(" ");
            else printf("\n");
        }
    }
    return 0;
}

1003 Colmeraner

预处理出不同区间的 $a*b$ 的值，然后求下每个点向四个方向能延伸的最大距离，求和即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;

typedef unsigned int ui;

int T;
int n,m;
int a[1100][1100];
ui l[1100][1100],r[1100][1100],u[1100][1100],d[1100][1100];
ui f[1100][1100]; 
int main()
{
    for (ui i=1;i<=1000;i++)
        for (ui j=1;j<=1000;j++)
            f[i][j]=f[i][j-1]+(i*(2*j+i-1)/2); 
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d %d",&n,&m);
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        memset(u,0,sizeof(u));
        memset(d,0,sizeof(d));
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
                scanf("%u",&a[i][j]);
        for (int i=1;i<=n;i++)  
        {
            l[i][1]=1;
            for (int j=2;j<=m;j++)
            {
                int now=j;
                while (now!=1&&a[i][j]<a[i][now-1]) now=l[i][now-1];
                l[i][j]=now;
            }
        }
        for (int i=1;i<=n;i++)
        {
            r[i][m]=m;
            for (int j=m-1;j>=1;j--)
            {
                int now=j;
                while (now!=m&&a[i][j]<a[i][now+1]) now=r[i][now+1];
                r[i][j]=now;
            }
        }
        for (int i=1;i<=m;i++)
        {
            u[1][i]=1;
            for (int j=2;j<=n;j++)
            {
                int now=j;
                while (now!=1&&a[j][i]>a[now-1][i]) now=u[now-1][i];
                u[j][i]=now;
            }
        }
        for (int i=1;i<=m;i++)
        {
            d[n][i]=n;
            for (int j=n-1;j>=1;j--)
            {
                int now=j;
                while (now!=n&&a[j][i]>a[now+1][i]) now=d[now+1][i];
                d[j][i]=now;
            }
        }
        ui ans=0;
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
                ans+=f[j-l[i][j]+1][r[i][j]-j+1]*f[i-u[i][j]+1][d[i][j]-i+1]*a[i][j];
        cout<<ans<<endl;
    }
    return 0;
}

1004 Dertouzos

答案应该为 $1$ 到 $min(\frac{n}{d},g)$ 的内的质数的个数，其中 $g$ 为 $d$ 的最小质因子。不难发现，由于这两个条件的制约，答案并不大，所以暴力枚举即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;
typedef long long LL;
#define maxn 2000010
bool valid[maxn];
int primm;
int prim[maxn];
void getPrime(int n)
{
    memset(valid,true,sizeof(valid));
    for (int i=2;i<=n;i++)
    {
        if (valid[i])
        {
            prim[++primm]=i;
            for (int j=2;j<=n/i;j++)
              valid[i*j]=0;
        }
    }
}
int n,d,g;
int main()
{
    getPrime(maxn-1);
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d %d",&n,&d);
        int i; 
        for (i=1;i<=primm;i++)
        {
            if (d*prim[i]>=n) break;
            if (d<prim[i]) break;
            if (d%prim[i]==0) break;
        }
        if (d*prim[i]>=n||d<prim[i]) i--;
        printf("%d\n",i);
    }
    return 0;
}

1005 Eade

先用ST表预处理数组，然后对于序列中的每个点二分出其能作为最大值的最大的区间 $[l,r]$ 。将所有区间排序，对于区间 $[l,r]$ 相同的点，其数值也一定相同，设其为 $x$ ，且出现的位置为 $p_1,p_2,\cdots,p_m$ 。构建新的数组 $c[0]=p_1-l+1,c[1]=p_2-p_1,\cdots,c[m]=r-p_m+1$ 。那么在这个区间内，最大值为 $x$ 的子区间对答案的贡献应为

z k = \sum i = 0 m c i * c i + k

$z_k=\sum _{i=0}^mc_i*c_{i+k}$
这一部分可以用FFT快速求得。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;
typedef long long LL;
const double PI = acos(-1.0);
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0){r = _r; i = _i;}
    complex operator +(const complex &b){return complex(r+b.r,i+b.i);}
    complex operator -(const complex &b){return complex(r-b.r,i-b.i);}
    complex operator *(const complex &b){return complex(r*b.r-i*b.i,r*b.i+i*b.r);}
}x1[400000],x2[400000];
void change(complex y[],int len)
{
    int i,j,k;
    for(i=1,j=len/2;i<len-1;i++)
    {
        if(i<j)swap(y[i],y[j]);
        k=len/2;
        while(j>=k)
        {
            j-=k;
            k/=2;
        }
        if(j<k) j+=k;
    }
}
void fft(complex y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        complex wn(cos(on*2*PI/h),sin(on*2*PI/h));
        for(int j = 0;j < len;j+=h)
        {
            complex w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                complex u = y[k], t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].r /= len;
}
int T,n;
int a[61000];
const int MAX=100000;
struct node
{
    int l,r,id;
}b[61000];
int cmp(node a,node b)
{
    if (a.l!=b.l) return a.l<b.l;
    else if (a.r!=b.r) return a.r<b.r;
    else return a.id<b.id;
}
int stTable[MAX][32];
int preLog2[MAX];

void st_prepare(int n,int *array)
{
    preLog2[1]=0;
    for (int i=2;i<=n;i++)
    {
        preLog2[i]=preLog2[i-1];
        if ((1<<preLog2[i]+1)==i)
        {
            preLog2[i]++;
        }
    }
    for (int i=n;i>=1;i--)
    {
        stTable[i][0]=array[i];
        for (int j=1;(i+(1<<j)-1)<=n;j++)
        {
            stTable[i][j]=max(stTable[i][j-1],stTable[i+(1<<j-1)][j-1]);
        }
    }
}
int query_max(int l,int r)
{
    int len=r-l+1,k=preLog2[len];
    return max(stTable[l][k],stTable[r-(1<<k)+1][k]);
}
int call(int x,int y)
{
    int l=0,r=y-1,ans;
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (query_max(y-mid,y)<=x) ans=mid,l=mid+1;
        else r=mid-1;
    }
    return y-ans;
}
int calr(int x,int y)
{
    int l=0,r=n-y,ans;
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (query_max(y,y+mid)<=x) ans=mid,l=mid+1;
        else r=mid-1;
    }
    return y+ans;
}
int num;
LL c[61000];
LL k[61000];
void work()
{
    int N=1;
    while (N<2*(num+1)) N*=2;
    for (int i=0;i<=num;i++)
        x1[i]=complex((double)c[i],0.0);
    for (int i=num+1;i<N;i++)
        x1[i]=complex(0.0,0.0);
    for (int i=0;i<=num;i++)
        x2[i]=complex((double)c[num-i],0.0);
    for (int i=num+1;i<N;i++)
        x2[i]=complex(0.0,0.0);
    fft(x1,N,1);
    fft(x2,N,1);
    for (int i=0;i<N;i++)
        x1[i]=x1[i]*x2[i];
    fft(x1,N,-1);
    for (int i=0;i<=num;i++)
    {
        LL d=(LL)(x1[num-i].r+0.5);
        k[i]+=d;
    }
}
int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        st_prepare(n,a);
        memset(k,0,sizeof(k));
        for (int i=1;i<=n;i++)
        {
            b[i].l=call(a[i],i);
            b[i].r=calr(a[i],i);
            b[i].id=i;
        }
        sort(b+1,b+1+n,cmp);
        num=1;
        c[0]=b[1].id-b[1].l+1;
        for (int i=2;i<=n;i++)
        {
            if (b[i].l==b[i-1].l&&b[i].r==b[i-1].r)
            {
                c[num++]=b[i].id-b[i-1].id;
            }
            else
            {
                c[num]=b[i-1].r-b[i-1].id+1;
                work();
                num=1;
                c[0]=b[i].id-b[i].l+1;
            }
        }
        c[num]=b[n].r-b[n].id+1; 
        work();
        LL ans=0;
        for (LL i=1;i<=n;i++)
            ans+=i^k[i];
        cout<<ans<<endl;
    }
    return 0;
}