Codeforces Round #669 ABC 题解

A. Ahahahahahahahaha

题意：给个一个偶数长度的01序列，要求删除不超过2/n个元素使得奇数位和等于偶数位和。

思路：注意到题目给的提示，只有0和1，且为偶数长度。
那么对和有贡献的也就只有1，而且0或1总有一个出现次数小于等于n/2。
所以我们就有这样的策略，把最后搞的只剩0或者1即可，0的个数小就删0，反之删1。
最后注意只保留1的时候还要考虑一下答案数组长度的奇偶性。

view code

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[maxn];
ll n;


int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        vector<ll> ans;
        n = read();
        rep(i,1,n) a[i] = read();
        ll sum = 0;
        rep(i,1,n) sum += a[i];
        if(sum > n/2)
        {
            rep(i,1,n) if(a[i]==1) ans.pb(a[i]);
            if(ans.size()&1)
            {
                ans.pop_back();
            }
        }
        else rep(i,1,n) if(a[i]==0) ans.pb(a[i]);
        if(ans.size()==1) ans[0] = 0;
        cout<<ans.size()<<'
';
        for(int i=0; i<ans.size(); i++) cout<<ans[i]<<' ';
        cout<<'
';
    }
    return 0;
}

B. Big Vova

题意：给一个a数组，问你构造一个b序列，使得存在c序列，其中c[i] = gcd(b[1],b[2],...,b[i])，且c的字典序最大。

思路：贪心，第一个肯定是最大的那个元素。然后每次暴力枚举出下一位能放的产生gcd最大的那个即可。

view code

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll n;
ll a[maxn];
ll vis[1005];


int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        vector<ll> ans;
        mem(vis,0);
        ll n = read();
        rep(i,1,n) a[i] = read();
        sort(a+1,a+1+n);
        ll m = a[n];
        ans.pb(m);
        ll cnt = 1;
        while(cnt<n)
        {
            ll ma = 0, pos = 0;
            per(i,n-1,1) if(!vis[i]&&gcd(a[i],m)>ma||(!vis[i]&&gcd(a[i],m)==ma&&a[pos]>a[i])) ma = gcd(a[i],m), pos = i;
            cnt++;
            ans.pb(a[pos]);
            m = ma;
            vis[pos] = 1;
    }
    rep(i,1,n) cout<<ans[i-1]<<' ';
    cout<<'
';
    }
    return 0;
}

C. Chocolate Bunny

题意：交互题，有一个n排列（没告诉你长什么样），你每次可以问它p[i]%p[j]的结果，然后他会告诉你这个值。让你在不超过2n的询问下把这个序列求出来。

思路：第一次做交互题刚开始有点蒙圈
其实想到一个点就可以立马A掉了。我们用双指针从头尾两端开始遍历。每次询问两个，(p,q)和(q,p)，因为肯定要么是a[p] > a[q] 要么是a[q] > a[p]，所以结果肯定一个等于两者之间的最小值，一个是比这个最小值要小。所以我们每两次询问都能得到一个a[p]或a[q]，然后移动指针即可。如2%5 和 5%2 ，分别是2和1，所以a[p]肯定是2。
具体详见代码。

view code

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[maxn];
ll vis[maxn];


int main()
{
    ll n = read();
    int p = 1, q = n;
    int cnt = 0;
    while(cnt<=n-1&&p<q)
    {
        printf("? %d %d
", p, q);
        fflush(stdout);
        ll r1 = read();
        printf("? %d %d
", q, p);
        fflush(stdout);
        ll r2 = read();
        if(r1 > r2) a[p] = r1, p++,vis[r1]=1;
        else a[q] = r2, q--, vis[r2]=1;
        cnt++;
    }
    ll sum = (1+n)*n/2;
    ll all = 0;
    rep(i,1,n) all += a[i];
    rep(i,1,n) if(a[i]==0) a[i] = sum - all;
    printf("! ");
    rep(i,1,n) printf("%lld%c",a[i], i==n?'
':' ');
    fflush(stdout);
    return 0;
}