upc组队赛3 Iranian ChamPions Cup

Iranian ChamPions Cup

题目描述

The Iranian ChamPions Cup (ICPC), the most prestigious football league in Iran, is reaching its end, and people are eagerly waiting for the finals, which happened to be between the two most popular Iranian teams, Persepolis and Esteghlal.
The ICPC finals consist of two matches, with each team competing as the home team in one match. The winning team is determined by aggregate score, the sum of the scores of the two matches. For example, if the scores of the two matches
are Persepolis 6–0 Esteghlal in the first match, and Esteghlal 3–1 Persepolis in the second match, then the aggregate score will be Persepolis 7–3 Esteghlal, meaning that Persepolis is the winner. If aggregates are equal, the away goals rule is used to determine the winner, in which case the winner is the team that scored the most goals in the match it played away from home. If the result is still equal, a penalty shootout is required.
Hana, an avid football fan, is trying to figure out various scenarios in which her favorite team wins the finals. To this end, she aims to write a program that gets as input the number of goals in the two matches, and decides which team is the winner if it can be derived from the aggregate scores and the away goals rule, otherwise declares that the match goes to penalty kicks. You are going to help Hana write such a program.

输入

The first line of the input contains two space-separated integers p1 and s1 , where p1 and s1 are the number of goals scored by Persepolis and Esteghlal, respectively, in the first match in which Persepolis is the home team. The second line contains two space-separated integers s2 and p2 , where s2 and p2 are the number of goals scored by Esteghlal and Persepolis, respectively, in the second match in which Esteghlal is the home team. All input integers are between 0 and 20, inclusively.

输出

In the output, print the name of the winning team, either Persepolis or Esteghlal, if the winner can be determined by the aggregate scores and the away goals rule. Otherwise, print Penalty.

样例输入

3 0
2 1

样例输出

Persepolis

题解

签到水题

代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef pair<int,int> P;
typedef long long LL;
typedef long long ll;
const double eps=1e-8;
const double PI = acos(1.0);
const int INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9+7;
const ll mod = 998244353;
const int MAXN = 1e6+7;
const int maxm = 1;
const int maxn = 100000+10;
int T;
int n,m;
int p1,p2;
int s1,s2;
int ans = 0;
int main(){
    read2(p1,s1);
    read2(s2,p2) ;
    int sum1 = p1 + p2;
    int sum2 = s1 + s2 ;
    if(sum1 < sum2) ans = 2;
    else if(sum2 < sum1) ans =1;
    else {
      if(p2 < s1) ans = 2;
      else if(p2 > s1) ans = 1;
      else ans = 3;
    }
    if(ans == 1)
    cout <<"Persepolis"<<endl;
    else if(ans == 2)
    cout <<"Esteghlal" <<endl;
    else
      cout <<"Penalty"<<endl;
}