10.16NOIP模拟赛

/*
我是一个大sb
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>

#define ll long long
#define N 10000007
#define mod 1000000000

using namespace std;
ll n,m,k,x,y,flag,ans,cnt;
ll a[N],cur[N];
priority_queue<ll>q;

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

int main()
{    
    freopen("shop.in","r",stdin);
    freopen("shop.out","w",stdout);
    n=read();m=read();
    x=read();y=read();
    a[1]=x;q.push(-a[1]);
    for(int i=2;i<=n;i++) 
    {
        a[i]=(y*a[i-1]+x)%mod;
        q.push(-a[i]);
        if(a[i]==0){k=i;break;}
    }
    if(!k)
    {
        for(int i=1;i<=m;i++)
        {
            ll res=q.top();q.pop();
            ans-=res;
        }
        cout<<ans<<endl;
        return 0;
    }
    int t=n/k,tt=n-t*k;a[t*k+1]=x;cur[++cnt]=x;
    for(int i=2;i<=tt;i++)
    {
        a[i+t*k]=(y*a[i+t*k-1]+x)%mod;
        cur[++cnt]=a[i];
    }
    sort(cur+1,cur+tt+1);
    if(m<=t) ans=0;
    else
    {
        while(!q.empty() && m)
        {
            int x=q.top();q.pop();x=-x;
            if(x<cur[1] || x>cur[tt]) 
            {
                m-=t;ans+=x*t;
            }
            if(x>=cur[1] && x<=cur[tt])
            {
                m-=(t+1);ans+=x*(t+1);
            }
        }
    }
    cout<<ans<<endl;
    fclose(stdin);fclose(stdout);
    return 0;
}

 

/*
优先队列
*/
#include<complex>
#include<cstdio>
#include<queue>
using namespace std;
const int mod=1e9;
long long n,m,x,y,ans;
priority_queue<int>q;
int main()
{
    freopen("shop.in","r",stdin);
    freopen("shop.out","w",stdout);
    scanf("%d%d%d%d",&n,&m,&x,&y);
    long long now=x;
    q.push(now);
    for(int i=2;i<=n;i++)
    {
        now=(now*y+x)%mod;
        if(i<=m)q.push(now);
        else if(now<q.top())
        {
            q.pop();
            q.push(now);
        }
    }
    while(!q.empty())
    {
        ans+=q.top();
        q.pop();
    }
    printf("%I64d\n",ans);
    fclose(stdin);fclose(stdout);
    return 0;
}

 

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007

using namespace std;
int n,m,k,cnt;
int a[N],b[N];
double c[N],ans;

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

void dfs(int now,int sum,double p)
{
    if(now==n+1)
    {
        ans+=double(sum)*p;
        return;
    }
    if(a[now]==0)
    {
         dfs(now+1,sum & b[now],p*(1.000-c[now]));
         dfs(now+1,sum,p*c[now]);
    }
    if(a[now]==1)
    {
         dfs(now+1,sum | b[now],p*(1.000-c[now]));
         dfs(now+1,sum,p*c[now]);
    }
    if(a[now]==2)
    {
        dfs(now+1,sum ^ b[now],p*(1.000-c[now]));
        dfs(now+1,sum,p*c[now]);
    } 
}

int main()
{
//    freopen("exp.in","r",stdin);
//    freopen("exp.out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++) a[i]=read();
    for(int i=1;i<=n;i++) b[i]=read();
    for(int i=1;i<=n;i++) scanf("%lf",&c[i]);
    ans=0;dfs(1,0,1);
    printf("%.1lf\n",ans);
    fclose(stdin);fclose(stdout);
    return 0;
}

 

/*
把每一位拆开单独考虑
f[i][0/1]表示到第i个运算，该位为0/1的概率
对于每一个运算符分情况讨论 
ans = sum[ f[n][1] * (1 << k) ]
*/
#include<bits/stdc++.h>

#define N 100007

using namespace std;
int n,a[N],b[N];
double c[N], f[N][2];

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

double solve(int bit)
{
    f[0][0]=1.00;
    for(int i=1;i<=n;i++)
    {
        int x = b[i] >> bit & 1;
        if(a[i]==0)
        {
            if(x==0)   f[i][0]=(f[i-1][1] + f[i-1][0]) * (1 - c[i]) + f[i-1][0] * c[i],
                       f[i][1]=f[i-1][1] * c[i];
            else        f[i][0]=f[i-1][0],
                       f[i][1]=f[i-1][1];
        }

        if(a[i]==1)
        {
            if(x==0)   f[i][0]=f[i-1][0],
                       f[i][1]=f[i-1][1];
            else        f[i][0]=f[i-1][0] * c[i],
                       f[i][1]=(f[i-1][0] + f[i-1][1]) * (1 - c[i]) + f[i-1][1] * c[i];
        }

        if(a[i] == 2)
        {
            if(x==0)   f[i][0]=f[i-1][0],
                       f[i][1]=f[i-1][1];
            else        f[i][0]=f[i-1][1] * (1 - c[i]) + f[i-1][0] * c[i],
                       f[i][1]=f[i-1][0] * (1 - c[i]) + f[i-1][1] * c[i];
        }
    }
    return f[n][1];
}
int main()
{
    freopen("exp.in", "r", stdin);
    freopen("exp.out", "w", stdout);
    cin>>n;
    for(int i = 1;i<=n;i++) cin>>a[i];
    for(int i = 1;i<=n;i++) cin>>b[i];
    for(int i = 1;i<=n;i++) scanf("%lf",&c[i]);
    double ans = 0;
    for(int i = 31;i>=0;i--) ans+=(double)(1ll << i) * solve(i);
    printf("%.1lf",ans);
    return 0;
}

 

、、

 

#include<bits/stdc++.h>

#define N 50001

using namespace std;
int n,m,T,ans,cnt,res,num;
int head[N],deep[N],sum[N];
int f[N][26],tmp1[N],tmp2[N];
struct edge{
    int u,v,w,net;
}e[N<<1];

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

inline void add(int u,int v,int w)
{
    e[++cnt].v=v;e[cnt].w=w;e[cnt].net=head[u];head[u]=cnt;
}

void dfs(int u,int fa,int c)
{
    f[u][0]=fa;deep[u]=c;
    for(int i=head[u];i;i=e[i].net)
    {
        int v=e[i].v;
        if(v==fa) continue;
        sum[v]+=e[i].w+sum[u];
        dfs(v,u,c+1);
    }
}

void get()
{
    for(int j=1;j<=25;j++)  for(int i=1;i<=n;i++)
    f[i][j]=f[f[i][j-1]][j-1];
}

int lca(int a,int b)
{
    if(deep[a]<deep[b]) swap(a,b);
    int t=deep[a]-deep[b];
    for(int i=0;i<=25;i++) 
    if(t&(1<<i)) a=f[a][i];
    if(a==b) return a;
    for(int i=25;i>=0;i--)
    {
        if(f[a][i]!=f[b][i]) 
        a=f[a][i],b=f[b][i];
    }return f[a][0];
}

void solve1(int x,int L,int k)
{
    cnt=0;
    while(x!=L)
    {
        tmp1[++cnt]=sum[x]-sum[f[x][0]];
        x=f[x][0];
    }
}
void solve2(int y,int L,int k)
{
    num=0;
    while(y!=L)
    {
        tmp2[++num]=sum[y]-sum[f[y][0]];
        y=f[y][0];
    }
} 

int main()
{
    //freopen("data.txt", "r", stdin);
    //freopen("2.out", "w", stdout);
    freopen("maze.in","r",stdin);
    freopen("maze.out","w",stdout);
    //freopen("ly.in","r",stdin);
    //freopen("ly.out","w",stdout);
    int x,y,z;
    n=read();
    for(int i=1;i<n;i++)
    {
        x=read();y=read();z=read();
        add(x,y,z);add(y,x,z);
    }
    dfs(1,1,1);get();
    m=read();
    for(int i=1;i<=m;i++)
    {
        x=read();y=read();z=read();
        int L=lca(x,y);
        memset(tmp1,0,sizeof tmp1);
        memset(tmp2,0,sizeof tmp2);
        solve1(x,L,z);
        solve2(y,L,z);reverse(tmp2+1,tmp2+num+1);
        ans=0;res=0;
        if(cnt)
        for(int i=1;i<=cnt;i++)
        {
            if(tmp1[i]<=res) res-=tmp1[i];
            else res=z,res-=tmp1[i],ans++;
        }
        if(num) 
        for(int i=1;i<=num;i++)
        {
            if(tmp2[i]<=res) res-=tmp2[i];
            else res=z,res-=tmp2[i],ans++;
        }
        printf("%d\n",ans);
    }
    fclose(stdin);fclose(stdout);
    return 0;
}

/*
LCA，压位
算 LCA 是显然的， 由于两个点之间可能要走很多步， 所以我们预处理每一个点往上连续
走六步会出现的各种情况，便能控制常数，从而在 8 秒内出解。由于边权只有两种，所以一
个点往上走六步遇到的边权只有 2^6 = 64 种情况
*/
#include <cstdio>
#include <cstring>
#define E 50010
#define F 100010
using namespace std;

int n, m, head[E], to[F], next[F], value[F], cnt = 0, cost = 0, costa = 0;

void Read(int &x)
{
    char ch = getchar();
    x = 0;
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
}

void add(int x, int y, int z)
{
    to[++cnt] = y; value[cnt] = z; next[cnt] = head[x]; head[x] = cnt;
}

int fa[E][20], dep[E], dist[E][7], type[E], jky[E][300], dd[E], jump[E];
short czy[1<<8][300][300][2];

void dfs(int u, int f, int d)
{
    fa[u][0] = f;
    dep[u] = d;
    for (int j = head[u]; j; j = next[j])
    {
        int v = to[j];
        if (v == f) continue;
        dist[v][0] = value[j];
        dfs(v, u, d+1);
    }
}

void build()
{
    for (int k = 1; k < 20; k++)
        for (int u = 1; u <= n; u++)
        {
            fa[u][k] = fa[fa[u][k-1]][k-1];
            if (k <= 5) dist[u][k] = dist[u][k-1] + dist[fa[u][k-1]][k-1];
        }
    
    for (int i = 1; i <= n; i++)
        if (dep[i] >= 6) dd[i] = dist[i][2] + dist[fa[i][2]][1];
        
    for (int i = 1; i <= n; i++)
    {
        if (dep[i] < 6) continue;
        int t = 0, u = i;
        for (int j = 0; j < 6; j++)
        {
            if (dist[u][0] == cost) t += 1 << j;
            u = fa[u][0];
        }
        jump[i] = u;
        type[i] = t;
    }

    for (int j = 0; j < (1<<6); j++)
        for (int k = cost; k <= cost * 6; k++)
            for (int rest = 0; rest <= k; rest++)
            {
                int path = j, step = 0, re = rest;
                for (int t = 1; t <= 6; t++)
                {
                    int co;
                    if (path & 1) co = cost; else co = costa;
                    if (re < co) re = k, step++;
                    re -= co;
                    path >>= 1;
                }
                czy[j][k][rest][0] = step;
                czy[j][k][rest][1] = re;
            }
        
    for (int i = 1; i <= n; i++)
    {
        int u = i, rest = 0;
        for (int j = 0; j < dd[i]; j++)
        {
            if (dep[u] > 0 && rest >= dist[u][0])
            {
                rest -= dist[u][0];
                u = fa[u][0];
            }
            jky[i][j] = u;
            rest++;
        }
    }
}

void swap(int& x, int& y)
{
    int t = x; x = y; y = t;
}


int get(int x, int y)
{
    if (dep[x] < dep[y]) swap(x, y);
    for (int k = 19; k >= 0; k--)
        if (dep[fa[x][k]] >= dep[y]) x = fa[x][k];
        
    if (x == y) return x;
    
    for (int k = 19; k >= 0; k--)
        if (fa[x][k] != fa[y][k]) x = fa[x][k], y = fa[y][k];
    if (x != y) x = fa[x][0];
    
    return x; 
}

struct hyd
{
    int step, rest;
};


inline hyd work(int u, int f, int k)
{
    int step = 0, rest = 0;
    
    if (k <= cost * 6)
        while (dep[u] > dep[f])
        {
            if (dep[u] - dep[f] < 6)
            {
                if (rest >= dist[u][0]) rest -= dist[u][0];
                    else step++, rest = k - dist[u][0];
                u = fa[u][0];
                continue;
            }
        
            step += czy[type[u]][k][rest][0];
            rest = czy[type[u]][k][rest][1];
            u = jump[u];

        }
    
    else
        while (dep[u] > dep[f])
        {
            if (rest == 0) step++, rest = k;
            
            if (dep[u] - dep[f] < 6)
            {
                if (rest >= dist[u][0]) rest -= dist[u][0];
                    else step++, rest = k - dist[u][0];
                u = fa[u][0];
                continue;
            }
            
            if (rest >= dd[u])
            {
                rest -= dd[u];
                u = jump[u];
                continue;
            }
        
            u = jky[u][rest], rest = 0;
        }
    
    return (hyd) {step, rest};
}

int main()
{
    freopen("data.txt", "r", stdin);
    freopen("1.out", "w", stdout);
    
    Read(n);
    memset(head, 0, sizeof(head));
    
    for (int i = 1; i < n; i++)
    {
        int x, y, z;
        Read(x), Read(y), Read(z);
        add(x, y, z);
        add(y, x, z);
        if (cost < z) cost = z;
        if (i == 1) costa = z; else if (costa > z) costa = z;
    }
    
    memset(fa, 0, sizeof(fa));
    memset(dep, 0, sizeof(dep));
    memset(dist, 0, sizeof(dist));
    
    dfs(1, 0, 0);
        
    build();
    
    Read(m);
    for (int i = 1; i <= m; i++)
    {
        int x, y, k;
        Read(x), Read(y), Read(k);
        int f = get(x, y);
        
        hyd ans1 = work(x, f, k), ans2 = work(y, f, k);
        int ans = ans1.step + ans2.step;
        ans -= (ans1.rest + ans2.rest) / k;
        
        printf("%d\n", ans);
    }
    
    fclose(stdin); fclose(stdout);
    
    return 0;
}