11.7NOIP模拟题

/*
有循环节
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define maxn 10000000
using namespace std;

int a,b,n,T;
int f[maxn];

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

int main()
{
    bool flag=false;
    a=read();b=read();n=read();
    f[]=;f[]=;f[]=(a+b)%;
    for(int i=;i<=n;i++)
    {
        f[i]=(a*f[i-]+b*f[i-])%;
        if(f[i-]==&&f[i]==)    {flag=true;T=i-;break;}
    }
    if(!flag)    printf("%d",f[n]);
    else
    {
        int pos=n%T;if(pos==)    pos=T;
        printf("%d",f[pos]);
    }
    return ;
}

/*
转化为二维平面上
左下角走到右上角不经过对角线的路径条数
卡特兰数列
*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 100000000
#define ll long long
#define RG register int
#define rep(i,a,b)    for(RG i=a;i<=b;i++)
#define per(i,a,b)    for(RG i=a;i>=b;i--)
using namespace std;
int T;
double n,m;
inline int read()
{
    int x=,f=;char c=getchar();
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

int main()
{
    T=read();
    while(T--)
    {
        n=read();m=read();
        if(n<m)    {puts("0.000000");continue;}
        double _1=n-m+,_2=n+;
        printf("%.6lf\n",_1/_2);
    }
    return ;
}

/*
Tarjan缩点后是一棵树
求树的直径并维护每个点到直径两端点的最大值。
三遍dfs即可
*/
#include<bits/stdc++.h>
#define maxn 20005
#define maxm 200005
using namespace std;

int n,m,id,dfn[maxn],low[maxn],head[maxn],head2[maxn],cnt;
int dis[maxn],dis1[maxn],mx=,root;
int belong[maxn],belnum;
bool vis[maxn];
stack<int> stk;
struct Edge{
int u,v,val,next;
}edge[maxm<<],e[maxm<<];

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

inline void add(int u,int v,int val)
{
    edge[++cnt].v=v;
    edge[cnt].u=u;
    edge[cnt].val=val;
    edge[cnt].next=head[u];
    head[u]=cnt;
} 

inline void tarjan(int u,int fa)
{
    dfn[u]=low[u]=++id;
    vis[u]=;
    stk.push(u);
    for(int i=head[u];i!=-;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            tarjan(v,u);
            low[u]=min(low[u],low[v]);
        }
        else if(vis[v]&&v!=fa)
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u])
    {
        belnum++;
        int temp;
        do{
            temp=stk.top();
            belong[temp]=belnum;
            vis[temp]=;
            stk.pop();
        }while(temp!=u);
    }
} 

inline void solve1()
{
    memset(head,-,sizeof(head));
    for(int i=,u,v,val;i<=m;i++)
    {
        u=read();v=read();val=read();
        add(u,v,val);add(v,u,val);
    }
    for(int i=;i<=n;i++)
    {
        if(!dfn[i])    tarjan(i,);
    }
} 

} 

namespace LP
{ 

inline void Add(int u,int v,int val)
{
    e[++cnt].v=v;
    e[cnt].val=val;
    e[cnt].next=head2[u];
    head2[u]=cnt;
} 

void dfs1(int u,int fa)
{
    for(int i=head2[u];i!=-;i=e[i].next)
    {
        int v=e[i].v;
        if(v==fa)    continue;
        dis[v]=dis[u]+e[i].val;
        if(dis[v]>mx)    mx=dis[v],root=v;
        dfs1(v,u);
    }
} 

void dfs2(int u,int fa)
{
    for(int i=head2[u];i!=-;i=e[i].next)
    {
        int v=e[i].v;
        if(v==fa)    continue;
        dis1[v]=dis1[u]+e[i].val;
        dfs2(v,u);
    }
} 

inline void solve2()
{
    cnt=;
    memset(head2,-,sizeof(head2));
    for(int i=;i<=n;i++)
        for(int j=head[i];j!=-;j=edge[j].next)
        {
            if(belong[i]!=belong[edge[j].v])
                Add(belong[i],belong[edge[j].v],edge[j].val);
        }
    dfs1(,-);
    mx=;memset(dis,,sizeof(dis));
    dfs1(root,-);
    mx=;
    dfs2(root,-);
    for(int i=;i<=n;i++)
        printf("%d\n",max(dis[belong[i]],dis1[belong[i]]));
} 

}
int main()
{
    freopen("prize.in","r",stdin);
    freopen("prize.out","w",stdout);
    n=read();m=read();
    Tarjan::solve1();
    LP::solve2();
    return ;
}