记录第一次Codeforces Round #250 (Div. 2)的比赛

今天好不容易赶到codeforces比赛的时间刚刚好，九点开始（之前一般是23:30开始，因为时差的原因），今天大概是在比赛开始后半小时才看的题，第一题很简单的一道题（基本上每次都有一道水题），今天就做了这一道题，还费了很大的劲，大概一个小时左右的时间，就有好几个人A掉了四题（一共五题），不得不羡慕他们的智商啊，先说一下我做的那道水题吧。

题意：

A. The Child and Homework

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices:A,B,C andD. Each choice has a description, and the child should find out the only one that is correct.

Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:

• If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
• If there is exactly one great choice then the child chooses it. Otherwise the child choosesC (the child think it is the luckiest choice).

You are given a multiple-choice questions, can you predict child's choose?

Input

The first line starts with "A." (without quotes), then followed the description of choiceA. The next three lines contains the descriptions of the other choices in the same format. They are given in order:B, C,D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.

Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".

Output

Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).

Sample test(s)

Input

A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute

Output

D

Input

A.ab
B.abcde
C.ab
D.abc

Output

C

Input

A.c
B.cc
C.c
D.c

Output

B

Note

In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choiceD (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will chooseD.

In the second sample, no choice is great, so the child will choose the luckiest choiceC.

In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will chooseB.

题目就是说小学生选选择题，如果有一个选项的长度小于其他三个选项的二倍，或者大于其他三个选项的二倍，则这个选项是最佳选项，否则选C。

我的代码比较繁琐，应该这题是属于简单题的，代码应该不是很长，不过博主太笨没有想到，只能非很大的劲来做了，等有空看看别的大神的代码，再更新下，下面放我的代码

(错误的貌似忽略掉了只有下划线的情况）：

#include 
#include
#include
using namespace std;
struct home
{
    int num;
    int length;
} len[4];
int cmp(const void *a,const void *b)
{
    home *p1=(home *)a,*p2=(home *)b;
    return p1->length-p2->length;
}//排序其实没有用在这，这是多余的
int main()
{
    char s[4][103];
    bool flag;
    int i,longer[4],shorter[4],j,k,l,con,coun;
    while(cin>>s[0])
    {
        len[0].length=strlen(s[0])-2;
        len[0].num=0;
        //l=0;
        coun=con=0;
        memset(longer,0,sizeof(longer));
        memset(shorter,0,sizeof(shorter));
        // k=0;
        for(i=1; i<4; i++)
        {
            cin>>s[i];
            len[i].length=strlen(s[i])-2;//记录长度
            len[i].num=i;
        }
        qsort(len,4,sizeof(len[0]),cmp);
        for(i=0; i<4; i++)
        {
            for(j=0; j<4; j++)
            {
                if(j!=i)
                {
                    if(len[i].length/len[j].length>=2)
                    {
                        longer[i]++;//记录是否大于其他选项的二倍
                        if(longer[i]==3)
                        {
                            k=i;
                            con++;
                        }
                    }
                }
            }
        }
        for(i=0; i<4; i++)
        {
            for(j=0; j<4; j++)
            {
                if(j!=i)
                {
                    if(len[j].length/len[i].length>=2)//&&(len[i].length/len[j].length!=1))
                    {
                        shorter[i]++;
                        if(shorter[i]==3)//记录是否少于其他选项的二倍
                        {
                            l=i;
                            coun++;
                        }
                    }
                }
            }
        }
        flag=false;
        for(i=0; i<4; i++)//输出
        {
            if((longer[i]==3&&con==1))
            {
                switch(len[k].num)
                {
                case 0:
                    cout<<"A"<

贴上一个大神的代码，果然精悍：

#include
#include
using namespace std;
int main(){
	string s;
	int i,a[4],j,tmp=0,k;
	bool b1,b2;
	for(i=0;i<4;++i){
		cin>>s;
		a[i]=s.length()-2;
	}
	for(i=0;i<4;++i){
		b1=true;b2=true;
		for(j=0;j<4;++j)if(j!=i){
			if(2*a[i]>a[j])b1=false;
			if(a[i]<2*a[j])b2=false;
		}
		if(b1||b2){tmp++;k=i;}
	}
	if(tmp==1)cout<<(char)('A'+k)<<'\n';
	else cout<<(char)'C'<<'\n';
	return 0;
}

这次比赛第一的代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

倍感自己太差啊，加油吧！！！难得cf比赛，还是挂零了。。。。