Codeforces Round #708 (Div. 2) ABC1C2题解

A. Meximization

第i位优先放等于i-1的，没有的话就后面随便填了。

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 endl '\n'
#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];
int vis[maxn];

int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        ll n = read();
        rep(i,1,n) a[i] = read(), vis[i] = 0;
        sort(a+1,a+1+n);
        int idx = 0;
        rep(i,1,n)
        {
            int flag = 0;
            rep(j,1,n) if(!vis[j]&&a[j]==i-1)
            {
                flag = 1;
                cout<<a[j]<<' ';
                vis[j] = 1;
                break;
            }

            if(!flag) break;
        }
        rep(i,1,n) if(!vis[i]) cout<<a[i]<<' ';
        cout<<endl;
    }
    return 0;
}

B. M-arrays

将a[i]存成模m后余数的形式，那么以每个a[i]与m-a[i]匹配的方式最优。特判一下a[i]为0和a[i]==m-a[i]的情况即可。

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 endl '\n'
#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];

int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        ll n = read(), m = read();
        rep(i,1,n) a[i] = read(), a[i] %= m;
        map<ll,ll> Map;
        map<ll,ll> vis;
        rep(i,1,n) Map[a[i]]++;
        ll ans = 0;
        rep(i,1,n)
        {
            if(vis[a[i]]) continue;
            if(a[i]==0) ans += 1;
            else
            {
                if(a[i]==m-a[i])
                {
                    ans += 1;
                }
                else
                {
                     ll mi = min(Map[a[i]], Map[m-a[i]]);
                    ll ma = max(Map[a[i]], Map[m-a[i]]);
                    ans += 1;
                    ans += max(ma-mi-1,0);
                }

            }
            vis[a[i]] = 1;
            vis[m-a[i]] = 1;
        }

        cout<<ans<<endl;
    }
    return 0;
}

k-LCM (hard version)

两个版本解法大同小异。先看C1，k=3时策略如下：

如果n为奇数，那么用a = (n-1)/2, b = a, c = 1即可。

若n为偶数，

且n/2是奇数，那么a = n/2, b = a, c = 2

否则直接 a = n/2, b = n/4, c = b即可。

那么C2的话，就先输出k-3个1，剩下的3个k来处理n-(k-3)又可以直接套C1方法了。

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 endl '\n'
#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} };

int main()
{
    int kase;
    cin>>kase;
    while(kase--)
    {
        ll n = read(), k = read();
        rep(i,4,k) cout<<1<<' ';
        n -= (k-3);
        ll a = 0, b = 0, c = 0;
        if(n&1)
        {
            a = n/2;
            b = a;
            c = 1;
        }
        else
        {
            if((n/2)&1)
            {
                a = n/2 - 1;
                b = a;
                c = 2;
            }
            else
            {
                a = n/2;
                c = a/2;
                b = c;
            }
        }
        cout<<a<<' '<<b<<' '<<c<<endl;
    }
    return 0;
}