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=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int main()
{
    bool flag=false;
    a=read();b=read();n=read();
    f[1]=1;f[2]=1;f[3]=(a+b)%7;
    for(int i=4;i<=n;i++)
    {
        f[i]=(a*f[i-1]+b*f[i-2])%7;
        if(f[i-1]==1&&f[i]==1)    {flag=true;T=i-2;break;}
    }
    if(!flag)    printf("%d",f[n]);
    else
    {
        int pos=n%T;if(pos==0)    pos=T;
        printf("%d",f[pos]);
    }
    return 0;
}

/*
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=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';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+1,_2=n+1;
        printf("%.6lf
",_1/_2);
    }
    return 0;
}

/*
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=0,root;
int belong[maxn],belnum;
bool vis[maxn];
stack<int> stk;
struct Edge{    
int u,v,val,next;
}edge[maxm<<1],e[maxm<<1];

inline 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*10+c-'0';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]=1;    
    stk.push(u);    
    for(int i=head[u];i!=-1;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]=0;    
            stk.pop();    
        }while(temp!=u);    
    }
} 

inline void solve1()
{    
    memset(head,-1,sizeof(head));    
    for(int i=1,u,v,val;i<=m;i++)    
    {    
        u=read();v=read();val=read();    
        add(u,v,val);add(v,u,val);    
    }    
    for(int i=1;i<=n;i++)    
    {    
        if(!dfn[i])    tarjan(i,0);    
    }
} 

} 

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!=-1;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!=-1;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=0;    
    memset(head2,-1,sizeof(head2));    
    for(int i=1;i<=n;i++)    
        for(int j=head[i];j!=-1;j=edge[j].next)    
        {    
            if(belong[i]!=belong[edge[j].v])    
                Add(belong[i],belong[edge[j].v],edge[j].val);    
        }    
    dfs1(1,-1);    
    mx=0;memset(dis,0,sizeof(dis));    
    dfs1(root,-1);    
    mx=0;    
    dfs2(root,-1);    
    for(int i=1;i<=n;i++)    
        printf("%d
",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 0;
}