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  • PAT (Advanced Level) 1036. Boys vs Girls (25)

    简单题。

    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<cstdio>
    #include<map>
    #include<queue>
    #include<vector>
    using namespace std;
    
    const int maxn=100000;
    struct X
    {
        string name;
        string sex;
        string id;
        int c;
    }s[maxn];
    int n;
    
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            cin>>s[i].name>>s[i].sex>>s[i].id>>s[i].c;
        int id1=-1,id2=-1;
        int Max=-1,Min=99999;
        for(int i=1;i<=n;i++)
            if(s[i].sex[0]=='F'&&s[i].c>Max) Max=s[i].c,id1=i;
    
        for(int i=1;i<=n;i++)
            if(s[i].sex[0]=='M'&&s[i].c<Min) Min=s[i].c,id2=i;
    
        if(id1==-1) printf("Absent
    ");
        else cout<<s[id1].name<<" "<<s[id1].id<<endl;
    
        if(id2==-1) printf("Absent
    ");
        else cout<<s[id2].name<<" "<<s[id2].id<<endl;
    
        if(id1==-1||id2==-1) printf("NA
    ");
        else printf("%d
    ",abs(s[id1].c-s[id2].c));
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5519545.html
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