Codeforces Round #563 (Div. 2)

CF

题目链接：https://codeforces.com/contest/1174

总结

ABC都比较水，27分钟写完了，但是D题想偏了，一直WA,AC代码确实太巧妙了。

Ａ题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
int n,sum1,sum2,a[N];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    read(n);
    for(int i=1;i<=2*n;i++)read(a[i]);
    sort(a+1,a+n*2+1);
    for(int i=1;i<=n;i++)sum1+=a[i];
    for(int i=n+1;i<=2*n;i++)sum2+=a[i];
    if (sum1==sum2){printf("-1");return 0;}
    for(int i=1;i<=2*n;i++)printf("%d ",a[i]);

Ｂ题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
int n,sum1,sum2,a[N],f[3];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    read(n);
    for(int i=1;i<=n;i++)read(a[i]),f[a[i]%2]++;
    if (f[0]>0&&f[1]>0)sort(a+1,a+n+1);
    for(int i=1;i<=n;i++)printf("%d ",a[i]);
}

C题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
int n,ans,a[N],f[N];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    read(n);
    for(int i=2;i<=n;i++)
    {
        if (f[i]==0)
        {
            ans++;
            f[i]=ans;
            int x=i;
            while(x+i<=n)f[x+i]=ans,x+=i;
        }
    }
    for(int i=2;i<=n;i++)printf("%d ",f[i]);
}

D题

题意：

一道构造题，给你n,x 构造尽量长的序列(｛a_l｝)，满足:

[egin{aligned} &ullet 1le a_ile 2^n \ &ullet 没有一个子序列的异或和等于零或x end{aligned} ]

题解

首先假设我们已经知道了这个序列，另s[i]表示i的亦或前缀和，那么对于任意一段子序列(i,j)都可以用s[i]^s[j]表示，

所以也就是不存在s[i]^s[j]=0或x。

所以我们可以先构造s[i]数组,不难发现一个性质：s[i]一定小于(2^n)，我们从(1->2^n)枚举,如果s[i]=k,那么k^x和k就不能再出现。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 1000050
int n,num,f[N],a[N];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    int x;
    read(n); read(x);
    int all=(1<<n)-1;
    f[x]=1;
    for(int i=1;i<=all;i++)if (!f[i]){a[++num]=i;f[i^x]=1;}
        
    printf("%d
",num);
    for(int i=1;i<=num;i++)
        printf("%d ",a[i]^a[i-1]);
}