Codeforces Round #672 (Div. 2)

身败名裂

A B C D

英语阅读题。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=5e4+11;
int T,N,A[MAXN];
int main(){
    T=read();
    while(T--){
        N=read();for(int i=1;i<=N;i++) A[i]=read();
        bool ff=1;for(int i=2;i<=N;i++) if(A[i-1]<=A[i]) ff=0;
        if(!ff) printf("YES\n");
        else printf("NO\n");
    }return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define int long long
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=1e5+11;
int T,N,A[MAXN],cnt[61],Ans;
signed main(){
    //freopen("1.in","r",stdin);
    T=read();
    while(T--){
        memset(cnt,0,sizeof(cnt));
        N=read();for(int i=1;i<=N;i++) A[i]=read();
        int ps=1;Ans=0;
        for(int i=1;i<=N;i++){
            int pw=1;
            for(int j=0;;j++){
                //cerr<<"j:"<<j<<" "<<pw<<endl;
                if(A[i]<pw){cnt[j-1]++;break;}
                pw*=2;
            }
        }
        for(int i=0;i<=60;i++) Ans+=(cnt[i]*(cnt[i]-1))/2;
        printf("%lld\n",Ans);
    }return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define int long long
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=3e5+11;
int T,N,d[MAXN],A[MAXN],Ans,Q;
void Modify(int ps,int w){
    if(d[ps]>=0) Ans-=d[ps];
    d[ps]=w-A[ps+1];
    if(d[ps]>=0) Ans+=d[ps];
    if(ps!=1){
        if(d[ps-1]>=0) Ans-=d[ps-1];
        d[ps-1]=A[ps-1]-w;
        if(d[ps-1]>=0) Ans+=d[ps-1];
    }
    A[ps]=w;return;
}
signed main(){
    //freopen("1.in","r",stdin);
    T=read();
    while(T--){
        N=read(),Q=read();for(int i=1;i<=N;i++) A[i]=read();A[N+1]=0;
        Ans=0;for(int i=1;i<=N;i++) d[i]=(A[i]-A[i+1]),Ans+=max(0ll,d[i]);
        printf("%lld\n",Ans);
        while(Q--){
            int l=read(),r=read();
            int w1=A[l],w2=A[r];
            Modify(l,w2);
            Modify(r,w1);
            printf("%lld\n",Ans);
        }
    }return 0;
}
 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define int long long
#define mod 998244353
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=6e5+11;
struct Segment{
    int Ans[MAXN<<2];
    void build(int k,int l,int r){
        Ans[k]=0;if(l==r) return;
        int mid=(l+r)>>1;build(k<<1,l,mid),build(k<<1|1,mid+1,r);
        return;
    }
    void Modify(int k,int l,int r,int ps){
        if(l==r){Ans[k]++;return;}
        int mid=(l+r)>>1;if(ps<=mid) Modify(k<<1,l,mid,ps);else Modify(k<<1|1,mid+1,r,ps);
        Ans[k]=Ans[k<<1]+Ans[k<<1|1];
    }
    int Query(int k,int l,int r,int x,int y){
        if(x<=l&&r<=y) return Ans[k];
        int mid=(l+r)>>1,res=0;
        if(x<=mid) res+=Query(k<<1,l,mid,x,y);
        if(mid<y) res+=Query(k<<1|1,mid+1,r,x,y);
        return res;
    }
}S;
int T,N,K,tmp[MAXN<<1],M,Ans,fac[MAXN],ifac[MAXN],inv[MAXN],L[MAXN],R[MAXN]; pii p[MAXN];
int C(int a,int b){if(a<b) return 0;return fac[a]*ifac[b]%mod*ifac[a-b]%mod;}
signed main(){
    //freopen("1.in","r",stdin);
    fac[0]=fac[1]=inv[1]=ifac[0]=ifac[1]=1;
    for(int i=2;i<MAXN;i++) fac[i]=fac[i-1]*i%mod,inv[i]=(mod-mod/i)*inv[mod%i]%mod,ifac[i]=ifac[i-1]*inv[i]%mod;
    T=1;
    while(T--){
        tmp[0]=0;
        N=read(),K=read();Ans=0;for(int i=1;i<=N;i++) L[i]=read(),R[i]=read(),tmp[++tmp[0]]=L[i],tmp[++tmp[0]]=R[i];
        sort(tmp+1,tmp+tmp[0]+1);M=unique(tmp+1,tmp+tmp[0]+1)-tmp-1;
        for(int i=1;i<=N;i++) p[i].fi=lower_bound(tmp+1,tmp+M+1,L[i])-tmp,p[i].se=lower_bound(tmp+1,tmp+M+1,R[i])-tmp;
        sort(p+1,p+N+1);S.build(1,1,M);
        for(int i=1;i<=N;i++){
            int res=S.Query(1,1,M,p[i].fi,M);
            Ans+=C(res,K-1);Ans%=mod;
            S.Modify(1,1,M,p[i].se);
        }printf("%lld\n",Ans);
    }return 0;
}
 

E

$O(n^5)$ 可过?

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<climits>
#include<bitset>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
const int MAXN=82;
int f[MAXN][MAXN][MAXN*MAXN],A[MAXN],P[MAXN],N,M,Ans,tot;
int calc(int x){return (x*(x-1))/2;}
int main(){
    //freopen("1.in","r",stdin);
    N=read();for(int i=1;i<=N;i++) A[i]=read();M=(N*(N-1))/2;
    for(int i=1;i<=N;i++) if(A[i]) P[++tot]=i;Ans=calc(N-tot);P[++tot]=N+1;
    memset(f,-127/3,sizeof(f));f[0][0][0]=0;
    for(int i=0;i<tot;i++){
        for(int j=i;j<=N+1;j++){
            for(int x=j+1;x<=N+1;x++){
                int dis=abs(P[i+1]-x),ww=calc(x-j-1);
                for(int pp=0;pp<=M;pp++) if(pp+dis<=M){
                    f[i+1][x][pp+dis]=max(f[i+1][x][pp+dis],f[i][j][pp]-ww);
                }
            }
        }
    }
    for(int i=0;i<=M;i++) {printf("%d ",Ans+f[tot][N+1][i]),f[tot][N+1][i+1]=max(f[tot][N+1][i+1],f[tot][N+1][i]);}
    printf("\n");
    return 0;
}

$O(n^4\log n)$ 的斜率优化做法 $dyls$ 的题解。 link

还有更加厉害的做法 link