2016大连网赛

网赛的时候就是前三个小时过了后面五道,然后两个小时就没搞出啥了;账号密码都忘了,只好重新写一遍了;

hdu-5868

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;

int vis[maxn],b[maxn],cnt=0,prime[maxn];
inline void Init()
{
    cnt=0;
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])
        {
            for(int j=2*i;j<maxn;j+=i)vis[j]=1;
            prime[++cnt]=i;
        }
    }
}
int phi(int x)
{
    int ans=x;
    for(int i=1;i<=cnt;i++)
    {
        if(x<prime[i])break;
        if(x%prime[i]==0)
        {
            ans=ans/prime[i]*(prime[i]-1);
            while(x%prime[i]==0)x/=prime[i];
        }
    }
    if(x>1)ans=ans/x*(x-1);
    return ans;
}
LL powmod(LL x,LL y)
{
    LL s=1,base=x;
    while(y)
    {
        if(y&1)s=s*base%mod;
        base=base*base%mod;
        y>>=1;
    }
    return s;
}
struct matrix
{
    LL a[2][2];
};
matrix cal(matrix A,matrix B)
{
    matrix C;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            C.a[i][j]=0;
            for(int k=0;k<2;k++)
            {
                C.a[i][j]+=A.a[i][k]*B.a[k][j];
                C.a[i][j]%=mod;
            }
        }
    }
    return C;
}
LL pow_mod(int y)
{
    matrix s,base;
    s.a[0][0]=s.a[1][1]=1;s.a[0][1]=s.a[1][0]=0;
    base.a[0][0]=base.a[0][1]=base.a[1][0]=1;base.a[1][1]=0;
    while(y)
    {
        if(y&1)s=cal(s,base);
        base=cal(base,base);
        y>>=1;
    }
    return (s.a[0][0]*3+s.a[0][1])%mod;
}
LL solve(int x)
{
    if(x==1)return 1;
    else if(x==2)return 3;
    else return pow_mod(x-2);
}
int main()
{
    int n;
    Init();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==1){printf("2
");continue;}
        int le=(int)sqrt(n*1.0+0.5),num=0;
        for(int i=1;i<=le;i++)
        {
            if(n%i==0)
            {
                b[++num]=n/i;
                if(b[num]*b[num]==n)continue;
                b[++num]=i;
            }
        }
        LL ans=0;
        for(int i=1;i<=num;i++)
        {
            ans=(ans+phi(n/b[i])*solve(b[i]))%mod;
        }
        ans=ans*powmod((LL)n,mod-2)%mod;
        print(ans);
    }
    return 0;
}

/*
题意：
https://async.icpc-camp.org/d/546-2016
*/

　　

hdu-5869

hdu-5870

hdu-5871

hdu-5872

hdu-5873

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e4+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;

int a[maxn];
int main()
{
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        For(i,1,m)
        {
            int n,flag=1;
            LL sum=0;
            scanf("%d",&n);
            For(i,1,n)scanf("%d",&a[i]);
            sort(a+1,a+n+1);
            for(int i=1;i<n;i++)
            {
                sum=sum+a[i];
                if(sum<(LL)i*(i-1))flag=0;
            }
            sum=sum+a[n];
            if(sum!=(LL)n*(n-1))flag=0;
            if(flag)printf("T
");
            else printf("F
");
        }
    }
    return 0;
}

/*
题意：
赢了得两分,平局得一分,输了的分,问得分情况是否符合实际;
思路：
排序后看是否满足这种游戏的要求;读入挂不知怎么T了;
*/

　　

hdu-5874

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e4+50;
const double eps=1e-8;


int main()
{
   LL n,m;
   while(cin>>n>>m)
   {
        LL a=n/2,b=(n+1)/2;
        if(a*b>m)printf("F
");
        else printf("T
");
   }
    return 0;
}

/*
题意：

*/

　　

hdu-5875

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=1e5+50;
const double eps=1e-8;

int n,m,a[maxn],d[maxn][23];

inline void rmq()
{
    for(int i=0;i<n;i++)d[i][0]=a[i];
    for(int j=1;(1<<j)<=n;j++)
        for(int i=0;i+(1<<j)<=n;i++)
            d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}
inline int query(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=R-L+1)k++;
    return min(d[L][k],d[R-(1<<k)+1][k]);
}

inline void solve(int u,int v)
{
    int temp=a[u];
    u++;
    while(u<=v)
    {
        int l=u,r=v;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            int t=query(u,mid);
            if(t>temp)l=mid+1;
            else r=mid-1;
        }
        if(l>v)break;
        if(a[l])temp%=a[l];
        if(temp==0)
        {
            printf("0
");
            return ;
        }
        u=l+1;
    }
    printf("%d
",temp);
}
int main()
{
    int t;
    read(t);
    while(t--)
    {
        read(n);
        for(int i=0;i<n;i++)read(a[i]);
        rmq();
        read(m);
        int l,r;
        while(m--)
        {
            read(l);read(r);
            l--;r--;
            solve(l,r);
        }
    }
    return 0;
}
/*
题意：
给一个数列,然后问[l,r]内这些数%得到多少;

思路：
比赛的时候队友用线段树A的,我现在写的是二分+rmq,就是让着里面的数递减才能改变它的结果;
*/

　　

hdu-5876

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e5+50;
const double eps=1e-8;

int n,m,ans[maxn];
pair<int,int>p;
map<pair<int,int>,int>mp;
queue<int>q,qu;
inline void makepair(int u,int v)
{
    p.first=u;
    p.second=v;
    mp[p]=1;
}
int main()
{
    int t;
    read(t);
    while(t--)
    {
        mp.clear();
        read(n);read(m);
        int u,v,s;
        For(i,1,m)
        {
            read(u);read(v);
            makepair(u,v);
            makepair(v,u);
        }
        read(s);
        while(!q.empty())q.pop();
        while(!qu.empty())qu.pop();
        For(i,1,n)
        {
            if(s==i)continue;
            qu.push(i);
        }
        q.push(s);
        ans[s]=0;
        while(!q.empty())
        {
            int fr=q.front();
            q.pop();
            int siz=qu.size();
            p.first=fr;
            while(siz--)
            {
                int f=qu.front();
                qu.pop();
                p.second=f;
                if(!mp[p])ans[f]=ans[fr]+1,q.push(f);
                else qu.push(f);
            }
        }
        int num=0;
        For(i,1,n)
        {
            if(i==s)continue;
            num++;
            if(num<n-1)printf("%d ",ans[i]);
            else printf("%d
",ans[i]);
        }
    }
    return 0;
}
/*
题意：
求一个点多边少的补图的一个起点到其他所有点的最短距离;

思路：
补图里面的边就是原图里面没有的边,那么每次就判断两个点之间在原图中是否有边,主要是原图里面边较少可以这样判断;
*/

　　

hdu-5877

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=2e5+50;
const double eps=1e-8;

int a[maxn],b[maxn],c[maxn],cnt,sum[maxn],in[maxn],n;
LL k,ans;
vector<int>ve[maxn];

int lowbit(int x){return x&(-x);}
inline int query(int x)
{
    int s=0;
    while(x)
    {
        s+=sum[x];
        x-=lowbit(x);
    }
    return s;
}
inline void update(int x,int num)
{
    while(x<=cnt)
    {
        sum[x]+=num;
        x+=lowbit(x);
    }
}
inline int getpos(LL x)
{
    int l=1,r=cnt;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if((LL)c[mid]>x)r=mid-1;
        else l=mid+1;
    }
    return r;
}

void dfs(int cur)
{
    int pos;
    if(a[cur]==0)pos=cnt;
    else pos=getpos(k/a[cur]);
    ans=ans+query(pos);

    int temp=getpos(a[cur]);
    update(temp,1);
    int len=ve[cur].size();
    for(int i=0;i<len;i++)dfs(ve[cur][i]);
    update(temp,-1);
}
int main()
{
    int t;
    read(t);
    while(t--)
    {
        read(n);read(k);
        For(i,0,n)
        {
            ve[i].clear();
            sum[i]=in[i]=0;
        }
        For(i,1,n)read(a[i]),b[i]=a[i];
        sort(b+1,b+n+1);
        c[1]=b[1];cnt=1;ans=0;
        For(i,2,n)
        {
            if(b[i]==b[i-1])continue;
            c[++cnt]=b[i];
        }
        int u,v,root;
        For(i,1,n-1)
        {
            read(u);read(v);
            ve[u].push_back(v);
            in[v]++;
        }
        For(i,1,n)
        {
            if(in[i])continue;
            root=i;
            break;
        }
        dfs(root);
        print(ans);
    }    
    return 0;
}
/*
题意：
求每个节点u以及和它的祖先节点v满足a[u]*a[v]<=k的对数和;

思路：
dfs的过程中将树状数组中保留的是它的所有祖先节点的信息,可以离散化后进行询问,
然后把这个点的信息更细到树状数组中,待它的所有节点都访问完了再删除,据说treap也可以做;
*/