第十一周 11.8-11.14

11.8

闲来无事补CF。

CF 592 C The Big Race

题目比较简单。点在于有一个大数比较可以用log弄掉。

算LCM要先除再乘。算log要先强转double。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;

LL gcd(LL a,LL b)
{
    return b?gcd(b,a%b):a;
}

int main(void)
{
    LL t,w,b;
    scanf("%I64d%I64d%I64d",&t,&w,&b);
    LL x,GCD=gcd(w,b),LCM=b/GCD*w;
    if(log(double(w))+log(double(b))>log(double(t))+log(double(GCD))) x=min(min(w-1,b-1),t);
    else x=t/LCM*min(w,b)+min(t%LCM,min(w-1,b-1));
    GCD=gcd(x,t);
    printf("%I64d/%I64d\n",x/GCD,t/GCD);
    return 0;
}

 

11.9

什么都没干。

 

11.10

CF 592 D Super M

在树上找包含标记节点的最小子树。dfs最远点对。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
# define maxn 150100
int cnt,headlist[maxn];
int is[maxn],val[maxn];
int d=0,A=maxn,B=maxn;

struct node
{
    int to,pre;
} edge[2*maxn];

void add(int from,int to)
{
    cnt++;
    edge[cnt].pre=headlist[from];
    edge[cnt].to=to;
    headlist[from]=cnt;
}

void dfs1(int pos,int fa,int dep)
{
    for(int i=headlist[pos];i;i=edge[i].pre)
    {
        int to=edge[i].to;
        if(to==fa) continue;
        dfs1(to,pos,dep+1);
        if(is[to]||val[to]) val[pos]+=2;
        val[pos]+=val[to];
    }
    if(is[pos]||val[pos])
    {
        if(dep>d) {d=dep; A=pos;}
        else if(dep==d) A=min(A,pos);
    }
    return;
}

void dfs2(int pos,int fa,int dep)
{
    if(dep>d) {d=dep; B=pos;}
    else if(dep==d) B=min(B,pos);
    for(int i=headlist[pos];i;i=edge[i].pre)
    {
        int to=edge[i].to;
        if(to==fa||!is[to]&&!val[to]) continue;
        dfs2(to,pos,dep+1);
    }
    return;
}

int main(void)
{
    int m,n;
    scanf("%d%d",&n,&m);
    for(int i=1;i<n;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v); add(v,u);
    }
    int s;
    for(int i=1;i<=m;i++)
    {
        int x; scanf("%d",&x);
        is[s=x]=1;
    }
    dfs1(s,0,0);
    d=0; dfs2(A,0,0);
    printf("%d\n%d\n",min(A,B),val[s]-d);
    return 0;
}

 

11.11-11.13

什么都没干。

 

11.14

补个BC。

HDU 5564 Clarke and digits

每次都是用到矩阵去搜板。这样不好- -

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const int n=77;
int a[77]={
    0,0,0,0,0,0,0,
    0,1,0,0,0,0,0,
    0,0,1,0,0,0,0,
    0,0,0,1,0,0,0,
    0,0,0,0,1,0,0,
    0,0,0,0,0,1,0,
    0,0,0,0,0,0,1,
    1,0,0,0,0,0,0,
    0,1,0,0,0,0,0,
    0,0,1,0,0,0,0,
    1,2,2,1,1,1,1
};

struct Mat {
    LL mat[77][77];
    Mat(){memset(mat,0,sizeof(mat));}
};

Mat operator * (Mat a, Mat b) {
    Mat c;
    int i, j, k;
    for(k = 0; k < n; ++k) {
        for(i = 0; i < n; ++i) {
            for(j = 0; j < n; ++j) {
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j] % mod ;
                c.mat[i][j] %= mod ;
            }
        }
    }
    return c;
}

Mat operator ^ (Mat a, int k) {
    Mat c;
    int i, j;
    for(i = 0; i < n; ++i)
        for(j = 0; j < n; ++j)
            c.mat[i][j] = (i == j);

    for(; k; k >>= 1) {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main(void)
{
    int T; scanf("%d",&T);
    while(T--)
    {
        int l,r,k;
        scanf("%d%d%d",&l,&r,&k);
        Mat t;
        for(int i=0;i<=6;i++) t.mat[70+i][70+i]++;
        for(int i=0;i<=9;i++)
        {
            for(int j=0;j<=6;j++)
            {
                for(int p=0;p<=9;p++)
                {
                    if(i+p==k) continue;
                    int q=(j*3%7+p)%7;
                    int pos1=7*i+j,pos2=7*p+q;
                    t.mat[pos2][pos1]++;
                    t.mat[70+q][pos1]++;
                }
            }
        }
        Mat L,R=t^(r-1);
        if(l!=1) L=t^(l-2);
        LL ans1=0,ans2=0;
        for(int i=0;i<n;i++)
        {
            ans1=(ans1+L.mat[70][i]*a[i]%mod)%mod;
            ans2=(ans2+R.mat[70][i]*a[i]%mod)%mod;
        }
        printf("%I64d\n",(ans2-ans1+mod)%mod);
    }
    return 0;
}