清北考前刷题day5早安

/*
C(n,k)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
 
#define ll long long
#define N 1000007
#define M 1000000007
 
using namespace std;
ll n,k,ans,cnt1,cnt0;
ll inv[N]={,},fac[N]={,},f[N]={,};
 
inline ll 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;
}
 
inline void init()
{
    for(int i=;i<=N;i++)
    {
        fac[i]=(fac[i-]%M*i%M)%M;
        f[i]=((M-M/i)%M*f[M%i])%M;
        inv[i]=(inv[i-]*f[i]%M)%M;
    }
}
 
inline ll C(ll a,ll b)
{
    return fac[a]%M*inv[b]%M*inv[a-b]%M;
}
 
int main()
{
    freopen("cube.in","r",stdin);
    freopen("cube.out","w",stdout);
    n=read();k=read();init();
    printf("%I64d\n",C(n,k));
    return ;
}

/*
做最大生成树时标记哪些边被加了进去。然后把这些边排序。
如果询问权值w的货物，就在边中查找小于w的最大值假设拍第k大
答案就是k+1。 复杂度O(qlogn)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm> 
 
#define N 100007
 
using namespace std;
int n,m,q,ans,cnt,num;
int head[N],fa[N];
struct edge{
    int u,v,w,net,pos;
    bool operator < (const edge &a) const{
            return w>a.w;
    }
}e[N<<],E[N],tmp[N<<];
 
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;
}
 
inline void add(int u,int v,int w)
{
    e[++cnt].u=u;e[cnt].v=v;e[cnt].w=w;
    e[cnt].net=head[u];head[u]=cnt;
}
 
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
 
void kruskal()
{
    int r1,r2;
    sort(E+,E+m+);
    for(int i=;i<=m;i++)
    {
        r1=find(E[i].u),r2=find(E[i].v);
        if(r1==r2)continue;
        fa[r1]=r2;num++;
        add(E[i].u,E[i].v,E[i].w);add(E[i].v,E[i].u,E[i].w);
        tmp[num].pos=num;tmp[num].u=E[i].u,tmp[num].v=E[i].v;tmp[num].w=E[i].w;
        if(num==n-) break;
    }
}
 
int serch(int x)
{
    int l=,r=n-,mid;
    while(l<=r)
    {
        mid=l+r>>;
        if(tmp[mid].w>=x) l=mid+;
        else ans=mid,r=mid-;
    }return num-ans+;
}
 
int main()
{
    freopen("warehouse.in","r",stdin);
    freopen("warehouse.out","w",stdout);
    int x,y,z,k;
    n=read();m=read();q=read();
    for(int i=;i<=n;i++) fa[i]=i;
    for(int i=;i<=m;i++)
    {
        E[i].u=read();E[i].v=read();E[i].w=read();
    }kruskal();
    for(int i=;i<=q;i++)
    {
        x=read();
        if(x<=e[cnt].w){printf("1\n");continue;}
        if(x>e[].w){printf("%d\n",n);continue;}
        k=serch(x);
        printf("%d\n",k+);
    }
    return ;
}