Codeforces Round #485 Div. 1 vp记

　　A：对每种商品多源bfs一下每个点到该商品的最近距离，对每个点sort一下取前s个即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define int long long
#define N 100010
int n,m,k,s,a[N],p[N],d[110][N],b[110],q[N],t;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void bfs(int k)
{
    int head=0,tail=0;memset(d[k],42,sizeof(d[k]));
    for (int i=1;i<=n;i++) if (a[i]==k) q[++tail]=i,d[k][i]=0;
    do
    {
        int x=q[++head];
        for (int i=p[x];i;i=edge[i].nxt)
        if (d[k][edge[i].to]>d[k][x])
        {
            d[k][edge[i].to]=d[k][x]+1;
            q[++tail]=edge[i].to;
        }
    }while (head<tail);
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    n=read(),m=read(),k=read(),s=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=m;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    for (int i=1;i<=k;i++) bfs(i);
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=k;j++) b[j]=d[j][i];
        sort(b+1,b+k+1);
        int ans=0;
        for (int j=1;j<=s;j++) ans+=b[j];
        printf("%d ",ans);
    }
    return 0;
}

　　B：看起来非常玄乎事实上只是说两人交换次数的奇偶性不同。考虑将原排列变成该排列的最小交换次数，也即n-置换的循环节，这个东西的奇偶性是与交换次数的奇偶性有关的，因为每次交换会合并或拆开置换的两个循环。或者可以发现逆序对的奇偶性也是与其相关的，因为显然交换只会对两位置之间的区间产生影响，如果之间的数比两位置上的数都小或都大显然没有影响，否则每有一个这样的数逆序对变化两个。同时两位置本身带来一个逆序对变化，这样总逆序对变化量肯定是奇数。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000010
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,a[N];
bool flag[N];
ll s;
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("b.in","r",stdin);
    freopen("b.out","w",stdout);
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    int c=n;
    for (int i=1;i<=n;i++)
    if (!flag[i])
    {
        c--;flag[i]=1;
        int x=a[i];while (!flag[x]) flag[x]=1,x=a[x];
    }
    if ((c&1))
    {
        if (n&1) cout<<"Petr";else cout<<"Um_nik";
    }
    else
    {
        if (n&1) cout<<"Um_nik";else cout<<"Petr";
    }
    return 0;
}

　　1.5h之后C还是没有思路，自闭到结束。

　　result：rank 154

　　C：刚的过程中最接近正解的想法是对0~2n-1建一些辅助点，对每个点将每一位挖掉与其连边，然后将辅助节点取反后连边，但这样显然不对。事实上稍加(?)修改就行了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N (1<<22)
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],b[N],ans;
bool flag[N<<1];
void dfs(int k)
{
    flag[k]=1;
    if (k>(1<<m))
    {
        if (!flag[a[k^(1<<m)]]) dfs(a[k^(1<<m)]);
    }
    else
    {
        for (int i=0;i<m;i++)
        if (!(k&(1<<i))&&!flag[k|(1<<i)]) dfs(k|(1<<i));
        if (b[k^(1<<m)-1]&&!flag[b[k^(1<<m)-1]+(1<<m)]) dfs(b[k^(1<<m)-1]+(1<<m));
    }
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
#endif
    m=read(),n=read();
    for (int i=1;i<=n;i++) b[a[i]=read()]=i;
    for (int i=(1<<m)+1;i<=(1<<m)+n;i++)
    if (!flag[i]) dfs(i),ans++;
    cout<<ans;
    return 0;
}

　　D：题意莫名其妙，实际上是设f(x)为x所有分解因子方式中最小的因子和，求f(x) (x>=n)的最小值。一个古老的结论是因子之和固定时尽量取3能使乘积最大，有剩余时补2，于是只需考虑这样的数。求对数可以大致求出答案范围。枚举每一种可能，高精度求出即可。这需要多项式快速幂。于是就被各种问题折腾了一天。有多惨我已经不想回顾了，最后也不知道怎么过的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N (1<<19)
#define M 1500010
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,a[M],r[N],len,ans=100000000;
ll f[N],c[N];
char s[M];
const double PI=3.14159265358979323846;
struct complex
{
    double x,y;
    complex operator +(const complex&a) const
    {
        return (complex){x+a.x,y+a.y};
    }
    complex operator -(const complex&a) const
    {
        return (complex){x-a.x,y-a.y};
    }
    complex operator *(const complex&a) const
    {
        return (complex){x*a.x-y*a.y,x*a.y+y*a.x};
    }
}g[N],b[N];
void DFT(complex *a,int n,int op)
{
    for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=2;i<=n;i<<=1)
    {
        complex wn=(complex){cos(2*PI/i),op*sin(2*PI/i)};
        for (int j=0;j<n;j+=i)
        {
            complex w=(complex){1,0};
            for (int k=j;k<j+(i>>1);k++,w=w*wn)
            {
                complex x=a[k],y=w*a[k+(i>>1)];
                a[k]=x+y,a[k+(i>>1)]=x-y;
            }
        }
    }
}
int work(int n,int c)
{
    c*=pow(3,n%6);n/=6;
    memset(f,0,sizeof(f));memset(b,0,sizeof(b));memset(g,0,sizeof(g));
    g[0].x=1;b[0].x=729;int t;
    for (t=2;t<=(n<<1);t<<=1)
    {
        for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
        DFT(b,t,1);
        if (n&(t>>1))
        {
            DFT(g,t,1);
            for (int i=0;i<t;i++) g[i]=g[i]*b[i];
            DFT(g,t,-1);
            for (int i=0;i<t;i++) f[i]=g[i].x/t+0.5;
            for (int i=0;i<t;i++) if (f[i]>999) f[i+1]+=f[i]/1000,f[i]%=1000;
            for (int i=0;i<t;i++) g[i].x=f[i],g[i].y=0;
        }
        if (t>n) break;
        for (int i=0;i<t;i++) b[i]=b[i]*b[i];
        DFT(b,t,-1);
        for (int i=0;i<t;i++) f[i]=b[i].x/t+0.5;
        for (int i=0;i<t;i++) if (f[i]>999) f[i+1]+=f[i]/1000,f[i]%=1000;
        for (int i=0;i<t;i++) b[i].x=f[i],b[i].y=0;
    }
    for (int i=0;i<t;i++) f[i]=g[i].x;
    for (int i=0;i<t;i++) f[i]=f[i]*c;
    for (int i=0;i<t;i++) if (f[i]>999) f[i+1]+=f[i]/1000,f[i]%=1000;
    while (f[t]==0) t--;return t+1;
    //c*729^n
}
int mul(ll *f,int c)
{
    int n=N-1;while (f[n]==0) n--;
    for (int i=0;i<=n;i++) f[i]*=c;
    for (int i=0;i<=n;i++) f[i+1]+=f[i]/1000,f[i]%=1000;
    if (f[n+1]) n++;
    return n+1;
}
bool check(ll *f,int len)
{
    int m=(n-1)/3+1;
    if (len>m) return 1;
    if (len<m) return 0;
    for (int i=m-1;~i;i--)
    if (f[i]!=a[i]) return f[i]>a[i];
    return 1;
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
#endif
    scanf("%s",s+1);n=strlen(s+1);
    for (int i=0;i<n;i++) a[i]=s[n-i]-‘0‘;
    if (n==1&&a[0]==1) {cout<<1;return 0;}
    for (int i=0;i<(n-1)/3+1;i++) a[i]=a[i*3]+a[i*3+1]*10+a[i*3+2]*100;
    len=work(max(floor((n-1)/log10(3))-1,0.0),1);
    for (int k=max(floor((n-1)/log10(3))-1,0.0);k<=ceil(n/log10(3))+1;k++)
    {
        memcpy(c,f,sizeof(c));
        if (check(c,len)) {ans=min(ans,k*3);break;}
        if (check(c,mul(c,2))) {ans=min(ans,k*3+2);break;}
        if (check(c,mul(c,2))) {ans=min(ans,k*3+4);}
        len=mul(f,3);
    }
    cout<<ans;
    return 0;
}

　　E：先考虑序列上怎么做。考虑每种质因子，问题变为求区间每个数和x取min之和。显然每个数的质因子数量不会很多，所以对每种质因子个数分别建个BIT即可，复杂度即为log2。可以线性筛预处理做到log分解质因数。树上问题直接树剖的话是log3，不太能接受，可以通过欧拉序去一个log，即在进入子树时加，退出子树时减。我又被，卡常了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 100010
#define M 10000000
#define P 1000000007
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,p[N],a[N],prime[M+10],minp[M+10],deep[N],fa[N][20],in[N],out[N],ans[N],tree[30][N<<1],cnt,m,t;
bool flag[M+10],tmp[30];
struct data{int to,nxt;
}edge[N<<1];
struct data2{int x,y,z,lca;
}q[N];
struct data3{int x,cnt;};
vector<data3> point[M+10],Q[M+10];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void add(int p,int k,int x){while (k<=2*n) tree[p][k]+=x,k+=k&-k;}
int query(int p,int k){int s=0;while (k) s+=tree[p][k],k-=k&-k;return s;}
int ksm(int a,int k)
{
    int s=1;
    for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
    return s;
}
void dfs(int k)
{
    in[k]=out[k]=++cnt;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k][0])
    {
        fa[edge[i].to][0]=k;
        deep[edge[i].to]=deep[k]+1;
        dfs(edge[i].to);
        out[k]=++cnt;
    }
}
int lca(int x,int y)
{
    if (deep[x]<deep[y]) swap(x,y);
    for (int j=19;~j;j--) if (deep[fa[x][j]]>=deep[y]) x=fa[x][j];
    if (x==y) return x;
    for (int j=19;~j;j--) if (fa[x][j]!=fa[y][j]) x=fa[x][j],y=fa[y][j];
    return fa[x][0];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("e.in","r",stdin);
    freopen("e.out","w",stdout);
#endif
    flag[1]=1;cnt=0;
    for (int i=2;i<=M;i++)
    {
        if (!flag[i]) prime[++cnt]=i,minp[i]=i;
        for (int j=1;j<=cnt&&prime[j]*i<=M;j++)
        {
            flag[prime[j]*i]=1;
            minp[prime[j]*i]=prime[j];
            if (i%prime[j]==0) break;
        }
    }
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    cnt=0;dfs(1);
    fa[1][0]=1;
    for (int j=1;j<20;j++)
        for (int i=1;i<=n;i++)
        fa[i][j]=fa[fa[i][j-1]][j-1];
    for (int i=1;i<=n;i++)
    {
        int x=a[i]=read();
        while (x>1)
        {
            int y=minp[x],cnt=0;
            while (minp[x]==y) cnt++,x/=y;
            point[y].push_back((data3){i,cnt});
        }
    }
    m=read();cnt=0;
    for (int i=1;i<=m;i++)
    {
        q[i].x=read(),q[i].y=read(),q[i].z=read();q[i].lca=lca(q[i].x,q[i].y);ans[i]=1;
        int x=q[i].z;
        while (x>1)
        {
            int y=minp[x],cnt=0;
            while (minp[x]==y) cnt++,x/=y;
            Q[y].push_back((data3){i,cnt});
        }
    }
    fa[1][0]=0;
    int u=25;
    for (int i=1;i<=M;i++)
    if (!flag[i])
    {
        while (ksm(i,u)>M) u--;
        for (int j=0;j<point[i].size();j++)
        {
            add(point[i][j].cnt,in[point[i][j].x],1);
            add(point[i][j].cnt,out[point[i][j].x]+1,-1);
            tmp[point[i][j].cnt]=1;
        }
        for (int j=0;j<Q[i].size();j++)
        {
            ll x=0;
            for (int k=1;k<=Q[i][j].cnt;k++)
            if (tmp[k]) x+=k*(query(k,in[q[Q[i][j].x].x])+query(k,in[q[Q[i][j].x].y])+
            1ll*(P-2)*query(k,in[q[Q[i][j].x].lca])%(P-1)+1ll*(P-2)*query(k,in[fa[q[Q[i][j].x].lca][0]])%(P-1));
            for (int k=Q[i][j].cnt+1;k<=u;k++)
            if (tmp[k]) x+=Q[i][j].cnt*(query(k,in[q[Q[i][j].x].x])+query(k,in[q[Q[i][j].x].y])+
            1ll*(P-2)*query(k,in[q[Q[i][j].x].lca])%(P-1)+1ll*(P-2)*query(k,in[fa[q[Q[i][j].x].lca][0]])%(P-1));
            ans[Q[i][j].x]=1ll*ans[Q[i][j].x]*ksm(i,((x%(P-1))+P-1)%(P-1))%P;
        }
        for (int j=0;j<point[i].size();j++)
        {
            add(point[i][j].cnt,in[point[i][j].x],-1);
            add(point[i][j].cnt,out[point[i][j].x]+1,1);
            tmp[point[i][j].cnt]=0;
        }
    }
    for (int i=1;i<=m;i++) printf("%d\n",ans[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/Gloid/p/10211755.html

时间： 2024-10-02 18:45:16

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