2020牛客暑期多校训练营（第三场）

A.Clam and Fish

记录答案为ans，很明显当为3、4的时候，捕鱼即可，ans++,当为0的时候，
判断是否有做好的鱼饵，有鱼饵的话直接钓鱼，ans++，此时鱼饵数也对应减一，
为1时，有蛤，故做一个鱼饵，鱼饵数+1。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <stack>
#include <cmath>
#include <stdlib.h>
#include <time.h>
#include <iomanip>
#define S(X) scanf("%d",&(X))
#define SS(X, Y) scanf("%d%d",&(X),&(Y))
#define SSS(X,Y,Z) scanf("%d%d%d",&(X),&(Y),&(Z))
#define SL(X) scanf("%lld",&(X))
#define SLL(X, Y) scanf("%lld%lld",&(X),&(Y))
using namespace std;
typedef double db;
typedef long long ll;
const int mod = 998244353;
const int maxn=2e6+100;
const double pi=acos(-1);
int main()
{
	#ifdef ONLINE_JUDGE
    #else
        freopen(".vscode/in.txt","r",stdin);
    #endif
    int t;
    S(t);
    while(t--)
    {
        int n;
        S(n);
        char s[maxn];
        scanf("%s",s);
        int er=0,ans=0,make=0;;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='0')
            {
                if(make)
                {
                    ans++;make--;
                }
            }
            else if(s[i]=='1')
            {
                    make++;
            }
            else 
                ans++;
           // cout<<"i="<<i<<"  er="<<er<<"  make="<<make<<"  ans="<<ans<<endl;
        }
        printf("%d\n",ans+make/2);
    }
	return 0;
}

B.Classical String Problem

这个直接将字符串首尾相连组成一个环即可，我们可以定义一个指针ops，这个指针就是更改后的字符串的初始位置，令初始值为0，当我们查询字符串x位置的字符时输出str[(ops+x-1+len)%len]即可，当x>0时，我们将指针顺时针走x个，即将一开始的字符串的初始位置+x，这样前x个就到字符串后面去了，这个位置就是更改后的字符串位置，因为可能会超过len的长度，因此要(ops+x+len)%len,当x为负的时候，同理ops也要加上x，因为x是负的，直接加即可，又因为可能结果小于0因此(ops+x+len)%len,这两个直接合并就好。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <stack>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define S(X) scanf("%d",&(X))
#define SS(X, Y) scanf("%d%d",&(X),&(Y))
#define SSS(X,Y,Z) scanf("%d%d%d",&(X),&(Y),&(Z))
#define SL(X) scanf("%lld",&(X))
#define SLL(X, Y) scanf("%lld%lld",&(X),&(Y))
using namespace std;
typedef double db;
typedef long long ll;
const int maxn=2e6+100;
const double eps=1e-5;
using namespace std;

int main()
{
    #ifdef ONLINE_JUDGE
    #else
        freopen(".vscode/in.txt","r",stdin);
    #endif
    char str[maxn];
    scanf("%s",str);
    getchar();
    int len=strlen(str);
    int q;
    S(q);
    char ch[5];
    int x;
    int ops=0;
    while(q--)
    {
        scanf("%s%d",ch,&x);
        //cout<<ch<<" "<<x<<endl;
        if(ch[0]=='A')
            {
                //cout<<"ops="<<ops<<" "<<(ops+x-1+len)%len<<endl;
                printf("%c\n",str[(ops+x-1+len)%len]);
            }
        else
        {
            //if(x>0)
                ops=(ops+x+len)%len;
            //else
               // ops=(ops-x+len)%len;
            //cout<<"ops="<<ops<<endl;
        }
        
    }
    return 0;
}

C.Operation Love

这题呢，找到长度为6和长度为9的两个相临边即可，记录这三个点组成的长度为6（端点为sp,op)和9（端点为op,ep)的两个边，然后按照sp,op,ep的顺序求叉乘即可，大于0就说明短边向长边逆时针旋转左手，小于0说明顺时针旋转右手。一开始设的eps为1e-8，然后改了各种方法结果发现是eps的问题，改了1e5就过了，血亏。。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <stack>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define S(X) scanf("%d",&(X))
#define SS(X, Y) scanf("%d%d",&(X),&(Y))
#define SSS(X,Y,Z) scanf("%d%d%d",&(X),&(Y),&(Z))
#define SL(X) scanf("%lld",&(X))
#define SLL(X, Y) scanf("%lld%lld",&(X),&(Y))
using namespace std;
typedef double db;
typedef long long ll;
const int maxn=2e5+100;
const double eps=1e-5;
using namespace std;
struct Point{
    long double x,y;
    Point(long double _x=0,long double _y=0){
        x=_x;
        y=_y;
    }
    Point operator-(const Point& p1)const{return Point(x-p1.x,y-p1.y);}
    Point operator+(const Point& p1)const{return Point(x+p1.x,y+p1.y);}
    bool friend operator<(const Point& p1,const Point& p2){
        if(fabs(p1.x-p2.x)>eps)return p1.x<p2.x;return p1.y<p2.y;
    }//map点的时候
    bool friend operator==(const Point& p1,const Point& p2){
        if(fabs(p1.x-p2.x)>eps)return 0;
        if(fabs(p1.y-p2.y)>eps)return 0;
        return 1;
    }
    long double friend operator^(const Point& p1,const Point& p2){return p1.x*p2.y-p2.x*p1.y;}//共线的时候
    Point friend operator*(const Point& p1,const double& k){return Point(p1.x*k,p1.y*k);}
    long double friend operator*(const Point& p1,const Point& p2){return p1.x*p2.x+p1.y*p2.y;}
};
long double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
long double Cross(Point &sp, Point &ep, Point &op)
{
    return ((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
long double Cross1(Point A, Point B){
    return (A.x*B.y-A.y*B.x);
}
long double Area(Point* p, int n){
    long double s = 0;
    for(int i = 1; i < n-1; ++i)
        s += Cross1(p[i]-p[0], p[i+1]-p[0]);
    return s;
}
int main()
{
    #ifdef ONLINE_JUDGE
    #else
        freopen(".vscode/in.txt","r",stdin);
    #endif
    int n;
    S(n);
    while(n--)
    {
        Point p[100];
        Point a,b,c,d;
        Point sp,op,ep;
        for(int i=0;i<20;i++)
        {
            scanf("%Lf%Lf",&p[i].x,&p[i].y);
            if(i!=0)
            {
                long double len=dis(p[i],p[i-1]);
                //cout<<"len="<<len<<endl;
                if(fabs(len-9.0)<eps)
                {
                    a=p[i];b=p[i-1];
                }
                else if(fabs(len-6.0)<eps)
                {
                    c=p[i];
                    d=p[i-1];
                }
            }
        }
        long double len1=dis(p[0],p[19]);
        // cout<<"len1="<<len1<<endl;
        if(fabs(len1-9.0)<eps)
        {
            //cout<<"haha"<<endl;
            a=p[0];
            b=p[19];
        }
        else if(fabs(len1-6.0)<eps)
        {
            //cout<<"haha"<<endl;
             c=p[0];
             d=p[19];
        }
        // cout<<"a= ";
        // cout<<a.x<<" "<<a.y<<endl;
        //  cout<<"b= ";
        // cout<<b.x<<" "<<b.y<<endl;
        //  cout<<"c= ";
        // cout<<c.x<<" "<<c.y<<endl;
        //  cout<<"d= ";
        // cout<<d.x<<" "<<d.y<<endl;
        if(a==c)
        {
            op=a;
            sp=d;
            ep=b;
        }
        else if(a==d)
        {
            op=a;
            sp=c;
            ep=b;
        }
        else if(b==c)
        {
            op=b;
            sp=d;
            ep=a;
        }
        else if(b==d)
        {
            op=b;
            sp=c;
            ep=a;
        }
         //cout<<sp.x<<" "<<sp.y<<endl;
         //cout<<op.x<<" "<<op.y<<endl;
        // cout<<ep.x<<" "<<ep.y<<endl;
        // cout<<"Cross=";
        // cout<<Cross(sp,ep,op)<<endl;
        // Point pp[4];
        // pp[0]=sp,pp[1]=op,pp[2]=ep;
        // cout<<"Area="<<Area(pp,3)<<endl;
        if(Cross(sp,ep,op)>0)
            printf("left\n");
        else
            printf("right\n");
         
    }
    return 0;
}

L. Problem L is the Only Lovely Problem

签到题，设定一个字符串ans="lovely",然后只要比较前6个字符即可，是否为ans[i]或者toupper(ans[i])即可。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <stack>
#include <cmath>
#include <stdlib.h>
#include <time.h>
#include <iomanip>
#define S(X) scanf("%d",&(X))
#define SS(X, Y) scanf("%d%d",&(X),&(Y))
#define SSS(X,Y,Z) scanf("%d%d%d",&(X),&(Y),&(Z))
#define SL(X) scanf("%lld",&(X))
#define SLL(X, Y) scanf("%lld%lld",&(X),&(Y))
using namespace std;
typedef double db;
typedef long long ll;
const int mod = 998244353;
const int maxn=5e5+100;
const double pi=acos(-1);
int main()
{
	#ifdef ONLINE_JUDGE
    #else
        freopen(".vscode/in.txt","r",stdin);
    #endif
    string s;
    cin>>s;
    int len=s.length();
    string ss="lovely";
    bool flag=false;
    for(int i=0;i<6;i++)
    {
        if((s[i]==ss[i])||(s[i]==toupper(ss[i])))
            continue;
        else
        {
            flag=true;
            break;
        }
    }
    if(flag)
        printf("ugly\n");
    else
    {
        printf("lovely\n");
    }
    
	return 0;
}

posted @ 2020-07-19 21:47 branna 阅读(210) 评论(0) 编辑收藏举报

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branna

2020牛客暑期多校训练营（第三场）

A.Clam and Fish

B.Classical String Problem

C.Operation Love

L. Problem L is the Only Lovely Problem

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