Codeforces Round #268 (Div. 2) 题解

A I Wanna Be the Guy

解题思路

直接用一个set，最后判断一下是否有n个数。

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1e6+5;int n;set<int> a; int main(){    int t,x,y;    cin >> n;    cin >> x;    rep(i,1,x)    {        cin >> t;        a.insert(t);    }    cin >> y;    rep(i,1,y)    {        cin >> t;        a.insert(t);    }    if(a.size()==n)    {        cout <<"I become the guy."<<endl;       }    else    {        cout <<"Oh, my keyboard!"<<endl;    }    return 0;   } 

B. Chat Online

解题思路

用一个数组标记固定的a,b，然后枚举k,暴力搜索c+k,d+k。

代码如下：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1e6+5;int p,q,l,r;int ff[20005];int a,b,c[20005],d[20005];int main(){    int re =0 ;    cin >> p>>q>>l>>r;    rep(i,1,p)    {        cin >> a>>b;         for(int j=a;j<=b;j++)        {            ff[j]++;        }    }     rep(i,1,q)    {        cin >> c[i]>>d[i];    }    for(int k=l;k<=r;k++)    {        int ct =0;        rep(i,1,q)        {            if(ct) break;            for(int j=c[i]+k;j<=d[i]+k;j++)            {                if(ff[j])                {                    ct=1;                    break;                }            }        }        if(ct) re++;    }    cout << re<<endl;    return 0;   } 

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C. 24 Game 简单构造

解题思路

首先判断奇偶：
奇数：可以知道5个数1，2，3，4，5可以组成24，然后其他有偶数个，分别相减得1。
偶数：可以知道4个数1，2，3，4可以组成24，然后其他的数有偶数个，分别相减得1。

代码如下：

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}typedef  vector<int> vi;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1e6+5;int main(){    int n;    cin >> n;    if(n<4)    {        cout << "NO"<<endl;        return 0;    }    cout << "YES"<<endl;    if(n%2)    {        int ct = (n-5)/2;        while(n>5)        {            cout << n << " - " <<n-1<<" = "<< 1<<endl;            n-=2;        }        cout << "3 * 5 = 15"<<endl;        cout << "2 * 4 = 8"<<endl;        cout << "8 + 15 = 23"<<endl;        cout << "23 + 1 = 24"<<endl;        for(int i=0;i<ct;i++)        {            cout << "24 * 1 = 24"<<endl;        }    }    else    {        int ct = (n-4)/2;        while(n>4)        {            cout << n << " - " <<n-1<<" = "<< 1<<endl;            n-=2;        }        cout << "3 * 4 = 12"<<endl;        cout << "12 * 2 = 24"<<endl;        cout << "24 * 1 = 24"<<endl;        for(int i=0;i<ct;i++)        {            cout << "24 * 1 = 24"<<endl;        }    }    return 0;   }