Codeforces Round #281 (Div. 2) （A、B、C、D题）_html/css_WEB-ITnose

Yesterday’s CF match was quite enjoyable, but the rating just didn’t increase and I didn’t seize the opportunity to increase the rating. .

I could have passed four questions, but failed. After re-evaluation, I knelt down on two questions. . Alas:-(

A. Vasya and Football

Idea: count 1 for yellow cards and 2 for red cards.

Output when the number exceeds 2 for the first time. >

Question link: A. Vasya and Football

AC code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <string>#include <cmath>using namespace std;struct node {	char name[25];	int a[105];}home, away;int main(){	for(int i=0; i<105; i++)	{		home.a[i] = 0;		away.a[i] = 0;	}	scanf("%s %s", home.name, away.name);	int n;	scanf("%d", &n);	while(n--)	{		int t, e;		char ch1[3], ch2[3];		scanf("%d %s %d %s", &t, ch1, &e, ch2);		if(ch1[0]=='h')		{			if(ch2[0]=='y')			{				home.a[e]++;				if(home.a[e]==2)printf("%s %d %d\n", home.name, e, t);			}			else if(ch2[0]=='r')			{				home.a[e]+=2;				if(home.a[e]==2||home.a[e]==3)printf("%s %d %d\n", home.name, e, t);			}		}		else 		{			if(ch2[0]=='y')			{				away.a[e]++;				if(away.a[e]==2)printf("%s %d %d\n", away.name, e, t);			}			else if(ch2[0]=='r')			{				away.a[e]+=2;				if(away.a[e]==2||away.a[e]==3 ) printf("%s %d %d\n", away.name, e, t);			}		}	}	return 0;}

B. Vasya and Wrestling

Idea: First use sum to determine whether the score is 0, sum>0 => first, sum< ;0 => second,

sum=0, then determine the dictionary order, if it is the same again, determine the last action. >Note that sum needs to be long. >

AC code:

C. Vasya and Basketball

Idea: Sort all the balls, first assign them all as 3, and then reduce it to 2, and then judge the MAX of the intermediate process

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <string>#include <cmath>using namespace std;int judge(int a[], int b[], int na, int nb){	int i, j;	for(i=0, j=0; i<na, j<nb; i++, j++)	{		if(a[i]>b[i])return 1;		else if(a[i]<b[i])return 0;	}	if(i==na&&j!=nb)return 0;	else if(j==nb&&i!=na)return 1;	else if(i==na&&j==nb)return 2;}int main(){	long long sum=0;	int n, a[200005], b[200005], na=0, nb=0;	scanf("%d", &n);	int t;	for(int i=0; i<n; i++)	{		scanf("%d", &t);		if(t>0)a[na++] = t;		else if(t<0)b[nb++] = -t;		sum+=t;	}	if(sum>0)printf("first\n");	else if(sum<0)printf("second\n");	else if(judge(a,b,na,nb)==1)printf("first\n");	else if(judge(a,b,na,nb)==0)printf("second\n");	else if(judge(a,b,na,nb)==2&&t>0)printf("first\n");	else if(judge(a,b,na,nb)==2&&t<0)printf("second\n");	return 0;}

There was a little confusion when doing the questions yesterday. 🎜>

Question link: C. Vasya and Basketball

AC Code:

D. Vasya and Chess

Thinking: It seems like this question is a bit confusing. Question link: D. Vasya and Chess

AC code:

#include <cstdio>#include <cstring>#include <iostream>  #include <queue>  #include <map>  #include <set>  #include <vector>  #include <algorithm>   using namespace std;    #define LL long long  #define INF 0xfffffffpair<int,bool> p[400010];int main(){    int n, m;    scanf("%d", &n);    for(int i=0; i<n; i++)	{        scanf("%d", &p[i].first);        p[i].second=1;    }    scanf("%d", &m);    for(int i=n; i<n+m; i++)	{        scanf("%d", &p[i].first);        p[i].second=0;    }    sort(p, p+n+m);    p[n+m].first=-1;    int as=n*3, bs=m*3, ansa, ansb;    int MAX = -0xfffffff;        for(int i=0; i<=n+m; i++)	{        if(i==0||p[i].first!=p[i-1].first)		{            if(as-bs>MAX)			{                MAX=as-bs;                ansa=as;                ansb=bs;            }        }        if(p[i].second==1) as--;		else bs--;    }    printf("%d:%d\n", ansa, ansb);    return 0;}