标签：复杂度   getch   水题   define   turn   组合数   printf   scan   +=   

A.count

本场比赛最难的题...

隔板法组合数容斥 xjb 搞搞就好了

//by Judge
#include<cstdio>
#include<iostream>
#define Rg register
#define fp(i,a,b) for(Rg int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(Rg int i=(a),I=(b)-1;i>I;--i)
#define ll long long
using namespace std;
const int mod=998244353;
const int M=1e7+3;
ll n; int m,k,mx,r,ans,fac[M],inv[M];
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline void Pls(int& x,int y){if((x+=y)>=mod)x-=mod;}
inline int qpow(int x,int p=mod-2){ int s=1;
    for(;p;p>>=1,x=mul(x,x)) if(p&1) s=mul(s,x); return s;
}
inline int C(ll n,int m){
    if(n<mx) return mul(fac[n],mul(inv[m],inv[n-m]));
    Rg int s=inv[m]; fp(i,0,m-1) s=mul(s,(n-i)%mod); return s;
}
inline int f(int n){ if(n<0) return 0; int res=0,tp;
    fp(i,0,k) tp=mul(C(k,i),C(n-i*(m-1)+k-1,k-1)), // 选 i 个要放零头的，乘上其放置方案 
        Pls(res,mul(i&1?mod-1:1,tp)); return res;
}
int main(){ scanf("%lld%d%d",&n,&m,&k);
    mx=k*m,r=n%m; if(!m) return !printf("0\n");
    fac[0]=1; fp(i,1,n) fac[i]=mul(fac[i-1],i);
    inv[n]=qpow(fac[n]); fd(i,n,1) inv[i-1]=mul(inv[i],i);
    fp(i,0,k) Pls(ans,mul(f(i*m+r-k),C((n-r)/m-i+k-1,k-1)));
    // 前面的 f ，求出零头放置方案（放到 k 个数上全都小于 m） 
    // 后面的 C ，组合数解隔板法 
    return !printf("%d\n",ans);
}

tree

很水的题 \(n^2\) 的复杂度，随便搜搜就好了

//by Judge
#include<cstdio>
#include<cstring>
#include<iostream>
#define Rg register
#define fp(i,a,b) for(Rg int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(Rg int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(Rg int i=head[u],v=e[i].to;i;v=e[i=e[i].nxt].to)
#define open(S) freopen(S".in","r",stdin),freopen(S".out","w",stdout)
#define ll long long
using namespace std;
const int mod=998244353;
const int M=1e5+3;
typedef int arr[M];
#ifndef Judge
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
#endif
char buf[1<<21],*p1=buf,*p2=buf;
inline int Max(int x,int y){return x>y?x:y;}
inline int Min(int x,int y){return x<y?x:y;}
inline int mul(int x,int y){return 1ll*x*y%mod;}
inline void Pls(int& x,int y){if((x+=y)>=mod)x-=mod;}
inline int read(){ int x=0,f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0'; return x*f;
} int n,x,y,root,ans,a[M]; int pat,head[M];
struct Edge{ int to,nxt; }e[M<<1];
inline void add(int u,int v){
    e[++pat]={v,head[u]},head[u]=pat;
    e[++pat]={u,head[v]},head[v]=pat;
}
void dfs(int u,int fa,int mn,int mx){
    if(u>=root) Pls(ans,mul(mn,mx));
    go(u) if(v^fa) dfs(v,u,Min(mn,a[v]),Max(mx,a[v]));
}
int main(){ n=read(); fp(i,1,n) a[i]=read();
    fp(i,2,n) x=read(),y=read(),add(x,y);
    fp(i,1,n) dfs(root=i,0,a[i],a[i]); return !printf("%d\n",ans);
}

distance

n·q ，水题，注意别炸精度就好了

//by Judge
#include<cstdio>
#include<cstring>
#include<iostream>
#define Rg register
#define fp(i,a,b) for(Rg int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(Rg int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(Rg int i=head[u],v=e[i].to;i;v=e[i=e[i].nxt].to)
#define ll long long
using namespace std;
const int M=2003;
typedef int arr[M];
#ifndef Judge
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
#endif
char buf[1<<21],*p1=buf,*p2=buf;
inline void cmax(ll& x,ll y){if(x<y)x=y;}
inline int Min(ll x,ll y){return x<y?x:y;}
inline int read(){ int x=0,f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0'; return x*f;
} char sr[1<<21],z[20];int CCF=-1,Z;
inline void Ot(){fwrite(sr,1,CCF+1,stdout),CCF=-1;}
inline void print(ll x,char chr='\n'){
    if(CCF>1<<20)Ot();if(x<0)sr[++CCF]=45,x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++CCF]=z[Z],--Z);sr[++CCF]=chr;
} int n,q,root; ll ans,dis[M][M];
int pat,head[M];
struct Edge{ int to,val,nxt; }e[M<<1];
inline void add(int u,int v,int w){
    e[++pat]={v,w,head[u]},head[u]=pat;
    e[++pat]={u,w,head[v]},head[v]=pat;
}
void dfs(int u,int fa,ll d){ dis[root][u]=d;
    go(u) if(v^fa) dfs(v,u,d+e[i].val);
}
int main(){ n=read(),q=read(); Rg int x,y,z;
    fp(i,2,n) x=read(),y=read(),z=read(),add(x,y,z);
    fp(i,1,n) dfs(root=i,0,0);
    while(q--){ x=read(),y=read(),ans=0;
        fp(i,1,n) cmax(ans,Min(dis[x][i],dis[y][i])); print(ans);
    } return Ot(),0;
}

CSP-S全国模拟赛第二场 【nan】

标签：复杂度   getch   水题   define   turn   组合数   printf   scan   +=   

原文地址：https://www.cnblogs.com/Judge/p/11713815.html