[cf]Round #529 (Div. 3)

A.

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

char s[65];

int main()
{
    int n;scanf("%d",&n);
    scanf("%s",s);
    int x=0;
    for(int i=0;i<n;i+=x){
        printf("%c",s[i]);
        x++;
    }
    puts("");
    return 0;
}

B.

#include <iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

int a[100005];

int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+1+n);
    int ans=min(a[n]-a[2],a[n-1]-a[1]);
    printf("%d
",ans);
    return 0;
}

C.

将n写成k个2的幂次之和，可以先写n个1，如果k>n就输出NO，否则将多出来的(n-k)个1分配到前k个1中

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int a[200005];

int qpow(int a,int b){
  int ret=1;
  while(b){
    if(b&1) ret=ret*a;
    a=a*a;
    b>>=1;
  }
  return ret;
}

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    if(k>n){
        printf("NO
");
        return 0;
    }
    int r=n-k,cur=1;
    while(r>0&&cur<=k){
        for(int i=1;;i++){
            int num=qpow(2,i);
            if(num-1>r){
                a[cur]+=num/2-1,r-=num/2-1;
                cur++;
                break;
            }
        }
    }
    if(r>0) puts("NO");
    else{
        puts("YES");
        for(int i=k;i>=1;i--){
            if(i!=k) printf(" ");
            printf("%d",a[i]+1);
        }
        puts("");
    }
    return 0;
}

D.

给定每个人的下个人和下下个人是谁，解这n个人的排列顺序；先把n个人排列的圈构建出来，再确定正方向

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

vector<int> edge[200005];
int vis[200005];
int ans[200005];

void dfs(int n,int u){
  int siz=edge[u].size();
  for(int i=0;i<siz;i++){
    int v=edge[u][i];
    if(!vis[v]){
        ans[n+1]=v;
        vis[v]=1;dfs(n+1,v);
    }
  }
}

int main()
{
    int n;scanf("%d",&n);
    int a,b;
    for(int i=1;i<=n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        if(i==1) {a=u,b=v; if(a>b) swap(a,b);}
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    ans[1]=1;
    vis[1]=1;
    dfs(1,1);
    if(((ans[2]<ans[3])&&(a==ans[2]&&b==ans[3]))||((ans[2]>ans[3])&&(a==ans[3]&&b==ans[2]))){
        for(int i=1;i<=n;i++){
            if(i!=1) printf(" ");
            printf("%d",ans[i]);
        }
        puts("");
    }
    else{
        for(int i=n;i>=1;i--){
            if(i!=n) printf(" ");
            printf("%d",ans[i]);
        }
        puts("");
    }
    return 0;
}